GIFT  OF 
Dr.   Horace   Ivie 


EDUCATION  DEFT 


- 


// 

-  *    -</&? 


/t       >F-f  t-*'n* 


• 


* 


ADAMS'S  NEW  ARITHMETIC— REVISED  EDITION, 


ARITHMETIC, 


IN   WHICH   THE 


PRINCIPLES  OF  OPERATING  BY  NUMBERS 


ARE 


ANALYTICALLY   EXPLAINED 


AND 


SYNTHETICALLY    APPLIED 


ILLUSTRATED  BY  COPIOUS  EXAMPLES 


DESIGNED  FOR  THE  USE  OF  SCHOOLS  AND  ACADEMIES. 


BY  DANIEL  ADAMS,   M.   D., 

AUTHOR   OF    THE    SCHOLAR'S   ARITHMETIC,   SCHOOL  GEOGRAPHY,    ETC 


BOSTON: 

PUBLISHED  BY  PHILLIPS  &  SAMPSON. 
1848. 


A3 


GIFT  OF 

I A  Q^r    ,^<S<S^     \    VJ 


Eitered  according  to  Act  of  Congress,  in  the  year  1848,  by 

DANIEL    ADAMS,    M.   D. 
In  the  Clerk's  Office  of  the  District  Court  of  the  District  of  New  Hampshire. 


EDUCATION  DEFT 


Stereotyped    by 
GEORGE     A.    CURTIS; 

NEW  ENGLAND  T  rPE  AND  STEREOTYPE  FOUNDRY 
BOSTON 


PEEFACE. 


THE  '<  Scho  ar's  Arithmetic,"  by  the  author  of  the  present  work,  was 
first  published  in  1801.  The  great  favor  with  which  it  was  received  is 
an  evidence  tliat  it  was  adapted  to  the  wants  of  schools  at  the  time. 

At  a  subsequent  period  the  analytic  method  of  instruction  was  applied 
to  arithmetic,  with  much  ingenuity  and  success,  by  our  late  lamented 
countryman,  WARREN  COLBURN.  This' was  the  great  improvement  in 
the  modern  method  of  teaching  arithmetic.  The  author  then  yielded  to 
the  solicitations  of  numerous  friends  of  education,  and  prepared  a  work 
combining  the  analytic  with  the  synthetic  method,  which  was  published 
ui  1827,  with  the  title  of  "Adams'  New  Arithmetic." 

Few  works  ever  issued  from  the  American  press  have  acquired  so 
great  popularity  as  the  "  New  Arithmetic."  It  is  almost  the  only  work 
on  arithmetic  used  in  extensive  sections  of  New  England.  It  has  been 
re-published  in  Canada,  and  adapted  to  the  currency  of  that  province. 
It  has  been  translated  into  the  language  of  Greece,  and  published  in 
that  country.  It  has  found  its  way  into  every  part  of  the  United  States. 
In  the  state  of  New  York,  for  example,  it  is  the  text-book  in  ninety- 
three  of  the  one  hundred  and  fifty -five  academies,  which  reported  to  the 
regents  of  the  University  in  1847.  And,  let  it  be  remarked,  it  has  se 
cured  this  extensive  circulation  solely  by  its  merits.  Teachers,  super 
intendents,  and  committees  have  adopted  it  because  they  have  found  it 
fitted  to  its  purpose,  not  because  hired  agents  have  made  unfair  repre 
sentations  of  its  merits,  and,  of  the  defects  of  other  works,  seconding 
their  arguments  by  liberal  pecuniary  offers  —  a  course  of  dealing  re 
cently  introduced,  as  unfair  as  it  is  injurious  to  the  cause  of  education. 
The  merits  of  the  "New  Arithmetic"  have  sustained  it  very  success 
fully  against  such  exertions.  Instances  are  indeed  known,  in  which  it 
has  been  thrown  out  of  schools  on  account  of  the  "  liberal  offers"  of  those 
interested  in  other  works,  but  has  subsequently  been  readopted  without 
any  efforts  from  its  publishers  or  author. 

The  "  New  Arithmetic "  was  the  pioneer  in  the  field  which  it  has 
occupied.  It  is  not  strange,  then,  that  teachers  should  find  defects  and 
deficiencies  in  it  which  they  would  desire  to  see  removed,  though  they 
might  not  think  that  they  would  be  profited  by  exchanging  it  for  any 
other  work.  The  repeated  calls  of  such  have  induced  the  author  to  un 
dertake  a  revision,  in  which  labor  he  would  present  acknowledgments 
to  numerous  friends  for  important  and  valuable  suggestions.  Mr.  J. 
HOMER  FRENCH,  of  Phelps,  N.  Y.,  well  known  as  a  teacher,  has  been 
engaged  with  the  author  in  this  revision,  and  has  rendered  important 
aid.  Mr.  W.  B.  BUNNELL,  also,  principal  of  Yates  Academy,  N.  Y., 
formerly  principal  of  an  academy  in  Vermont,  has  assisted  throughout 
the  work,  having  prepared  many  of  the  articles.  The  revision  after 
Percentage  is  mostly  his  work. 


IV  PREFACE. 

The  characteristics  of  the  "  New  Arithmetic,"  which  have  given  the 
work  so  great  popularity,  are  too  well  known  to  require  any  notice  here. 
These,  it  is  believed,  will  be  found  in  the  new  work  in  an  improved 
form. 

One  of  the  peculiar  characteristics  of  the  new  work  is  a  more  natural 
and  philosophical  arrangement.  After  the  consideration  of  simple  whole 
numbers,  that  of  simple  fractional  numbers  should  evidently  be  intro 
duced,  since  a  part  of  a  thing  needs  to  be  considered  quite  as  frequently 
as  a  whole  thing.  Again ;  since  the  money  unit  of  the  federal  currency 
is  divided  decimally,  Federal  Money  certainly  ought  not  to  precede 
Decimal  Fractions.  It  has  been  thought  best  to  consider  it  in  connec 
tion  with  decimals.  Then  follow  Compound  Numbers,  both  integral 
and  fractional,  the  reductions  preceding  the  other  operations,  as  they 
necessarily  must.  Percentage  is  made  a  general  subject,  under  which 
are  embraced  many  particulars.  The  articles  on  Proportion,  Alligation, 
and  the  Progressions  will  be  found  well  calculated  to  make  pupils  thor 
oughly  acquainted  with  these  interesting  but  difficult  subjects. 

Care  has  been  taken  to  avoid  an  arbitrary  arrangement,  whereby  the 
processes  will  be  pureiy  mechanical  to  the  learner.  If,  for  instance,  all 
the  reductions  in  common  fractions  precede  the  other  operations,  the 
pupil  will  have  occasion  to  divide  one  fraction  by  another  long  before  he 
shall  have  learned  the  method  of  doing  it,  and  must  proceed  by  a  rule, 
to  himself  perfectly  unintelligible.  The  studied  aim  has  been  through 
out  the  entire  work  to  enable  the  ordinary  pupil  to  understand  every 
thing  as  he  advances.  The  author  is  yet  to  be  convinced  that  mental 
discipline  will  be  promoted,  or  any  desirable  end  be  subserved,  by  con 
ducting  the  pupil  through  blind,  mechanical  processes.  Just  so  far  as 
he  can  understand,  and  no  farther,  is  there  prospect  of  benefit.  No 
good  results  from  presenting  things,  however  excellent  in  themselves, 
if  they  are  beyond  the  comprehension  of  the  learner. 

Those  teachers  who  prefer  to  examine  their  classes  by  questions,  will 
find  that  little  will  escape  the  pupil's  attention,  who  shall  correctly  an 
swer  all  those  in  the  present  work,  while  teachers  who  practise  the  far 
superior  method  of  recitation  by  analysis,  will  find  the  work  admirably 
adapted  to  their  purpose. 

The  examples,  it  is  hoped,  will  require  very  full  applications  of  the 
principles. 

Many  antiquated  things,  which  it  has  been  fashionable  to  copy  in 
arithmetics,  from  time  immemorial,  have  been  omitted  or  improved, 
while  new  and  practical  matter  has  been  introduced.  A  Key  to  this 
revision  is  in  progress. 

With  these  remarks,  the  work  is  submitted  to  the  candid  examination 
of  the  public,  by  THE  AUTHOR. 

Keene,  N.  U.,  February.  1848. 


SUGGESTIONS  TO  TEACHERS. 

THE  writer  complies  with  the  request  of  the  venerable  author  of  '•'  Ad 
ams'  Arithmetic,''  to  preface  the  new  work  with  a  few  suggestions  lo 
his  associates  in  the  work  of  instruction.  Though  he  has  been  engaged 
for  sometime  past  in  assisting  to  make  the  work  better  fitted  to  Ac 
complish  its  design,  he  is  perfectly  satisfied  that  improvement  in  school 
education  is  rather  to  be  sought  in  improved  use  of  the  books  which  we 
now  have,  than  in  making  better  books.  Better  arithmeticians  would 
be  made  by  the  book  as  it  was  before  the  present  revision,  using  it  as 
it  might  be  used,  than  will  probably  be  made  in  most  cases  with  the  new 
work,  even  though  the  former  were  very  defective,  the  latter  perfect. 
Exertion,  then,  to  bring  teachers  to  a  higher  standard,  will  be  more 
effective  in  improving  school  education,  than  any  efforts  at  improving 
school  books  can  possibly  be.  It  is  here  where  the  great  improvement  in  ust 
be  sought.  Without  the  cooperation  of  competent  teachers,  the  greaiest 
excellences  in  any  book  will  remain  unnoticed,  and  unimproved.  Pu 
pils  will  frequently  complain  that  they  have  never  found  one  that  could 
explain  some  particular  thing,  of  which  a  full  explanation  is  givren  in 
the  book  which  they  have  ever  used,  and  their  attention  only  needed  to 
have  been  called  to  the  explanation. 

Then  let  teachers  make  themselves,  in  the  first  place,  thoroughly 
acquainted  with  arithmetic.  The  idea  that  they  can  "  study  and  keep 
ahead  of  their  classes/'  is  an  absurd  one.  They  must  have  surveyed 
the  whole  field  in  order  to  conduct  inquirers  over  any  part,  or  there  will 
be  liability  to  ruinous  misdirection.  Young  teachers  are  little  aware  of 
their  deficiencies  in  knowledge,  and  still  less  aware  of  the  injurious 
effect^  which  these  deficiencies  exert  upon  pupils,  who  are  often  dis 
gusted  with  school  education,  because  they  are  made  to  see  in  it  so  little 
that  is  meaning. 

In  the  next  place,  let  no  previous  familiarity  with  the  subject  excuse 
teachers  from  carefully  preparing  each  lesson  before  meeiing  their 
classes.  Thereby  alone  will  they  feel  that  freshness  of  interesi,  which 
will  awaken  a  kindred  interest  among  their  pupiis  5  and  if  on  any  occa 
sion  they  are  compelled  to  omit  such  preparation, -they  will  discover  a 
declining  of  interest  with  their  classes.  Teachers  who  are  obliged  to 
have  their  books  open,  and  watch  the  page  while  their  classes  recite,  are 
unfit  for  their  work. 

Pupils  should  be  taught  how  to  study.  That,  after  all,  is  the  great 
object  of  educating.  The  facilities  for  merely  acquiring  knowledge  are 
abundant,  if  persons  know  how  to  improve  them.  The  members  of 
classes  will  often  fail  in  recitation,  not  because  they  have  not  tried,  but 
have  not  known  how  to  get  their  lesson.  They  neglect  trying,  because 
they  can  do  so  little  to  advantage.  They  may  read  over  a  statement  in 
their  book  a  dozen  times,  they  say,  but  cannot  remember  it. — because 
they  do  not  understand  it.  An  hour  spent  with  each  pupil  individually 


VI  SUGGESTIONS    TO    TEACHERS. 

in  questioning  him  on  the  meaning  of  each  sentence,  which  ne  may 
be  required  to  read,  will  be  of  incalculable  advantage. 

When  pupils  shall  have  been  taught  how  to  study,  let  them  be  re 
quired  to  get  their  lessons,  and  recite  thtm.  If  the  present  book  is  not 
thought  by  teachers  to  contain  a  sufficient  description,  and  a  sufficient 
explanation  of  everything,  let  them  try  to  find  one  that  does,  for  if  pupils 
present  themselves  before  the  blackboard  at  the  time  of  recitation,  with 
the  expectation  that  the  teacher  is  to  explain  to  the  class,  and  help  them 
through  with  what  they  cannot  go  through  themselves,  they  will  not 
feel  that  they  must  have  studied  themselves ;  and  the  paltry  oralizing  of 
the  teacher  will  not  be  listened  to,  or  if  heard,  will  not  be  understood 
or  at  best,  not  retained  in  memory.  Pupils  may  be  made  to  see  things 
for  the  moment,  while  no  abiding  impression  will  remain  on  their 
minds.  They  will  often  proceed,  in  such  a  manner,  through  a  book, 
and  perhaps  have  the  mistaken  idea  that  they  understand  its  contents 

—  to  perpetuate   the   evil   of  superficialism,   perhaps,    themselves,  as 
teachers.     Pupils  will  never  have  a  sufficient  understanding  of  a  sub 
ject  till  they  shall  have  studied  it  carefully  themselves,  and  mastered 
each  part  by  severe  personal  application. 

Recitation  by  analysis  will  be  found  more  conducive  to  thorougn 
scholarship  than  adherence  to  any  written  questions.  Let  the  class,  or 
any  member  ot  the  class,  be  able  to  commence  at  the  beginning  and  go 
through  with  the  entire  lesson  without  any  suggestion  from  the  teacher, 

—  a  thing  that  is  perfectly  practicable  and  easily  attainable.    Let  pupils 
be  called  on,  at  the  pleasure  of  the  teacher,  in  any  part  of  the  class,  to 
go  on  with  the  recitation,  even  to  proceed  with  it  in  the  midst  of  a  sub 
ject,  the  topic  in  no  case  ever  being  named  by  the  teacher.     They  will 
thereby  become  accustomed  to  give  their  attention  to  the  recitation,  and 
they  will  be  profited  from  it,  besides  securing  habits  of  attention,  which 
will  be  of  incalculable  value. 

In  fine,  let  arithmetic  be  studied  properly,  and  more  valuable  mental 
discipline  will  be  acquired  from  it,  than  is  often  attained  from  the  whole 
course  in  mathematics  usually  assigned  by  college  faculties.  It  is  not 
the  extent,  but  the  value  of  acquisitions  in  mathematics,  which  is 
desirable.  W.  B.  B. 


INDEX. 


SIMPLE  NUMBERS. 

N  Cation  and  Numeration, 9    Contractions  in  Division, 59 

Addition 16     Review  of  Division, 63 

Review  of  Numeration  and  Addition,     .22    Miscellaneous  Exercises, 65 

Subtraction, 23  Problems  in  the  Measurement  of  Rec- 

Review  of  Subtraction, 29        tangles  and  Solids, 69 

Multiplication,     31 ,  illustration  by  Diagram,  .  71 

,  illustration  by  Diagram,  34     Definitions, 73 

Contractions  in  Multiplication,    .   .   .   .40    General  Principles  of  Division 74 

Review  of  Multiplication, 45     Cancelation, 75 

Division, 47    Common  Divisor, 78 

— ,  illustration  by  Diagram,  ...  50    Greatest  Common  Divisor, 78 

COMMON  FRACTIONS. 

Notation  of  Common  Fractions,  ....  80  Multiplication  of  whole  numbers  by  a 

Proper,  Improper,  £c., 82        fraction, 95 

Reduction  of  Fractions,      83  Multiplication    of    one     fraction    by 

To    reduce  a  fraction    to    its   lowest  another 9i 

terms,     85    General  Rule,      97 

Addition  and  Subtraction  of  Fractions,  .  87     Examples  in  Cancelation, 93 

Common  Denominator,  . 87    Division  of  Fractions, 99 

• ,  1  st  method.    .   .  S3 by  a  whole  num- 

,  2d  method,'    .   .  89        ber,  two  ways, 100 

Least  Common  Denominator,  or  Least  Division  of  whole  numbers  by  a  fraction,  101 

Common  Multiple, 90  Division  of  one  fraction  by  another,    .  103 

New  Numerators,' 91     General  Rule, 103 

General  Rule, 91  Reduction  of  Complex  to  Simple  Frac- 

Multiplication  of  Fractions,      93        tions, 104 

. by  a  whole  Promiscuous  Examples, 10' 

number,  two  ways, 94  Review  of  Common  Fractions,  ....  106 

DECIMAL  FRACTIONS  AND  FEDERAL  MONEY. 

Decimal  Fractions, 108  Addition  and  Subtraction  of  Decimal 

Notation  of  Decimal  Fractions,     ...  110        Fractions, 119 

Table, Ill  Addition   and  Subtraction  of  Federal 

To  read  Decimals 112        Money 120 

To  write  Decimals,    .  -. 112  Multiplication  of  Decimal  Fractions,      121 

Reduction  of  Decimal  Fractions,  .   .   .  113    ,  illustration  by  Diagram,  122 

. of  Common  to  Decimal  Frac-  of  Federal  Money,     .    .123 

tions, 114  Division  of  Decimal  Fractions,      .   .   .  124 

Federal  Money 116 of  Federal  Money, 126 

Reduction  of  Federal  Money,     .   .   .   .  118  Review  of  Decimal  Fractions,     .   .   .   .  127 

!  Bills, 129 

COMPOUND  NUMBERS. 

Definition, 131  MEASURE  OF  EXTENSION. 

Reduction  of  Compound  Numbers,  .   .  132  Reduction  of,  I.  Linear  Measure,     .   .  13S 

English  Money,    ....  132 Cloth  Measure,      ....  132 

WEIGHT.  II.  Land,      or       Square 

1.  Avoirdupoise  Weight,    135  Measure, 140 

II.  Troy  Weight,  ....  136    III.  Cubic  Measure,     .  .  141 

—  III.  Apothecary's  weight,  137 


INDEX. 


MEASURE  OP  CAPACITY. 

Reduction  of,  I.  Wine  Measure,   .       .143 

II.  Beer  Measure,    .   .   .144 

III.  Dry  Measure,    ...  144 

Time, 145 

Circular  Measure,     .   .   .  146 

Miscellaneous  Table, 147 

Reduction    of    Fractional    Compound 

Numbers, 147 

To  reduce   a  fraction  of  a  higher  de 
nomination  to  one  of  a  lower,     .   .   .  148 
To  reduce  a  fraction  of  a  lower  to  a 

higher  denomination 148 

To  reduce  a  fraction  of  a  higher  to  in- 

teuers  of  a  lower  denomination.     .    .  149 
To  reduce  integers  of  a  lower  to  frac 
tions  of  a  higher  denomination,     .    .  149 
Reduction  of  Decimal  Compound  Num 
bers 151 

Review    of    Reduction  of   Compound 

Numbers 153 

Addition  of  Compound  Numbers,      .    .  156 


Compound  mimners,      .    .  I 
Fractional  Compound  Num- 


bers, 160 

Subtraction  of  Compound  Numbers,     .  160 


Subtraction  of  Fractional  Compound 
Numbers,  164 

Multiplication  and  Division  of  Com 
pound  Numbers, 165 

Difference  in  longitude  and  time  be 
tween  different  places.  Diagram,  .  .  171 

Review  of  Compound  Numbers,    .   .   .  172 

Analysis,      174 

Given,  price  of  unity,  the  quantity,  to 
find  the  price  of  quantity, 175 

Given,  quantity,  price  of  quantity,  to 
find  the  price  of  unity, 

Given,  price  of  unity,  price  of  quantity, 
to  find  the  quantity, 

Practice.     Aliquot  Parts, 

Articles  sold  by  100, 

by  the  ton  of  2000  Ibs.,  .  .  . 

price,  aliquot  part  of  a  pound, 


175 

175 

178 
ISO 
181 


<fcc.. 


Articles,  quantity  less  than  unity,  .  .184 
To  reduce  shillings,  pence,  fee.,  to  the 

decimal  of  a  pound, 185 

To  reduce  the  decimal  of  a  pound  to 

shillings,  pence  and  farthings,    .   .   .  187 


PERCENTAGE. 


Definition,  187.     Rule, 188 

Insurance, 190 

Mutual  Insurance 191 

Stocks 193 

Brokerage, 194 

Profit  and  Loss,      194 

Interest,  195.     General  Rule, 199 

Easy  '.vay  of  castin?  interest  when  the 

rate  is  6  per  cent."     200 

To  compute  interest  on  pounds,  shil 
lings,  pence.  &c., 205 

To  compute  interest  when  partial  pay 
ments  have  been  made, 205 

Compound  Interest 209 

• Table, .211 

Annual  Interest, 212 

Time,  rate,  and  amount  given,  to  find 
the  principal 214 


Discount,  21 C.     Commission,     .   .   .   .  216 
Time,  rate,  interest,  to  find  the  princi 
pal,     217 

Principal,  interest,  time,  to  find  the  rate,  218 
Principal,   rate,   interest,  to  find    the 

lime,     218 

Percentage  to  find  the  rate, 219 

Bankruptcy, 220 

General  Average, 221 

Partnership,        222 

on  Time, 223 

Banking, 224 

Taxes,  method  of  assessing, 225 

Duties,     22? 

Specific 228 

Ad  Valorem, 229 

Review  of  Percentage,      230 

Equation  of  Payments, 233 


Inverted  and  Direct  Ratios,     

235 
236 
236 
237 
233 
239 

239 
240 
242 
244 
245 
246 
247 
251 

Practical  Exercises,      
Extraction  of  the  Cube  Root,      .   .   ,   . 
Practical  Exercises,       ..... 
Review 

266 
269 
873 
274 

275 
277 

Sff9 

281 
2-3 
2x-, 
288 

880 

liiK) 

291 
2:12 
293 
2!U 
2SS 

yoo 

361 
303 
304 

Rule  of  Three,   

To  invert  both  Ratios,  

Arithmetical  Progression  

To  find  the  fourth  term  of  a  proportion 
when  three  are  given,  

Simple  Interest  by  Progression,     .   .   . 
Annuities  by  Arithmetical  Progression, 

Cancelation  Applied      .   . 

Compound  Interest  by  Progression,  .    . 
Compound  Discount.  —  Table.  
Annuities  at  Compound  Interest,   .   .   . 
Present  worth  of  Annuities  at  Corn- 

Compound  Proportion  

Review  of  Proportion  .   .   . 

Alligation  Medial,     

Present  worth  of  Annuities.    Table,  . 
•  in  Rever- 

Exchange,    

Value  of  Gold  Coins,     
Duodecimals,      

254 
255 
256 

259 
260 
262 

Permutation,  

;  ;  ,  scale  for  taking  Dimen 
sions  in  feet  and  Decimals  of  a  foot, 

Measurement  of  Surfaces    

Evolution,  

Forms  of  Notes,  &c  .  . 
Billa,    .... 

ABITHMETJC. 


NOTATION    AND    NUMERATION. 

5T  1.  A  single  thing,  as  a  dollar,  a  horse,  a  man,  &c.,  is 
called  a  unit,  or  one.  One  and  one  more  are  called  two,  two 
and  one  more  are  called  three,  and  so  on.  Words  expressing 
how  many  (as  one,  two,  three,  &c.)  are  called  numbers. 

This  way  of  expressing  numbers  by  words  would  be  very 
slow  and  tedious  in  doing  business.  Hence  two  shorter  meth 
ods  have  been  devised.  Of  these,  one  is  called  the  Roman* 
method,  by  letters;  thus,  I  represents  oqe  ;  V,  five;  X,  ten, 
&c.,  as  shown  in  the  note  at  the  bottom  of  the  page. 

The  other  is  called  the  Arabic  method,  by  certain  charaC' 
ters,  called  figures.  This  is  that  in  general  use. 

*  In  the  Roman  method,  by  letters,  I  represents  one;  V,Jive;  X,  ten;  L, 
Jifly;  C,  one  hundred;  D,Jive  hundred;  and  M,  one  thousand. 

As  often  as  any  letter  is  repeated,  so  many  times  its  value  is  repeated,  un 
less  it  be  a  letter  representing  a  less  number  placed  before  one  representing  a 
greater;  then  the  less  number  is  taken  from  trie  greater  ;  thus,  IV  represents 
Jour;  IX,  nine,  &c.,  as  will  be  seen  in  the  following 

One 

Two 

Three 

Four 

Five 

Six 

Seven 

Eight 

Nine 

Ten 

Twenty 

Thirty 

Forty 

Fifty 

Sixty 

Seventy 

Eighty 

*  ID  is  used  instead  of  D  to  represent  five  hundred,  and  for  every  additional  0  an 
uexed  at  the  right  hand,  the  number  is  increased  ten  times. 

t  CtO  is  used  to  represent  one  thousand,  and  for  every  C  and  D  put  at  each  end,  the 
number  is  increased  ten  times. 
I  A  line  over  any  number  increases  its  value  one  thousand  times. 


TABLE. 

I. 

Ninety 

LXXXX.  or  XC 

II. 

One  hundred 

C. 

III. 

Two  hundred 

cc. 

IIII.  or  IV. 

Three  hundred 

COG. 

V. 

Four  hundred 

cccc. 

VI. 

Five  hundred 

D.  or  10.* 

VII. 

Six  hundred 

DC. 

VIII. 

Seven  hundred 

DCC. 

VIIII.  or  IX. 

Eight  hundred 

DCCC, 

X. 

Nine  hundred 

DCCCC. 

XX. 

One  thousand 

M.  or  ClO.t 

XXX. 

Five  thousand 

IOO.  or  V.t 

XXXX.  or  XL. 

Ten  thousand 

CCIOO.  or  X. 

L. 

Fifty  thousand 

IOOD. 

LX. 

Hundred  thousand 

CCCIOOO.  or 

C 

LXX. 

One  million 

M. 

LXXX. 

Two  million 

MM. 

10  NOTATION  AND  NUMERATION.  H  2,  3. 

In  the  Arabic  method  the  first  nine  numbers  have  each  a 
separate  character  to  represent  it ;  thus, 

V  ^3»     A   unit,   or,  single     »         --          m, 
Al_s™  •  .,-4  <L  •  xu-  Notel.   These  nine  char- 

thing,  is .  repMSaWjws  ;by  this     *      acterg  are  ca]}ed  significant 

character,  .     .     .     .     .     4     .  1.  figures,  because  they  each 

k.T>VQ..iifcitSo  uy.wijss  character,  2.  represent     some      number. 

Three  Units,  b^tni^'charciGter,  3.  Sometimes,   also,  they  are 

Four  units,  by  this  character,  4.  called  digits. 

Five  units,  by  this  character,  5.        Note  2-  '  The   value   of 

Six  units,  by  this  character,  6.  these  figures,  as  here  shown, 

0  V  Ji      .1  •  «  is  cal  ed  their  simple  value. 
Seven  units,  by  this  character,  7.  It  is  their  yalue  al/     when 

.bight  units,  by  this  character,     8.    smgie. 
Nine  units,  by  this  character,      9. 

Nine  is  the  largest  number  which  can  be  expressed  by  a 
single  figure.  There  is  another  character,  0 ;  it  is  called  a 
cipher,  naught,  or  nothing,  because  it  denotes  the  absence  of  a 
thing.  Still  it  is  of  frequent  use  in  expressing  numbers. 

By  these  ten  characters,  variously  combined,  any  number 
may  be  expressed. 

The  unit  1  is  but  a  single  one,  and  in  this  sense  it  is  called 
a  unit  of  the  first  order.  All  numbers  expressed  by  one  fig 
ure  are  units  of  the  first  order. 

IT  3.  Ten  has  no  appropriate  character  to  represent  it,  but 
it  is  considered  as  forming  a  unit  of  a  second  or  higher  order, 
consisting  of  tens.  It  is  represented  by  the  same  unit  figure 

1  as  is  a  single  thing,  but  it  is  written  at  the  left  hand  of  a 
cipher;  thus,  10,  ten.    The  0  fills  ihejlrst  place,  at  the  right 
hand,  which  is  the  place  of  units,  and  the  1  the  second  place 
from  the  right  hand,  which  is  the  place  of  tens.     Being  put 
in  a  new  place,  it  has  a  new  value,  which  is  ten  times  its 
simple  value,  and  this  is  what  is  called  a  local  value. 

Questions.  —  IT  1.  What  is  a  single  thing  called?  What  is  a 
uumoer  ?  Give  some  examples.  How  many  ways  of  expressing  num- 
oers  shorter  than  writing  them  out  in  words  ?  What  are  they  called  ? 
Which  is  the  method  in  general  use  ?  In  the  Arabic  method,  how  many 
numbers  have  each  a  separate  character? 

Tf  2.  How  is  one  represented  ?  Make  the  characters  to  nine.  What 
are  these  nine  characters  called  ?  Why  ?  What  is  the  simple  value  of 
Sgures?  What  is  the  largest  number  which  can  be  represented  by  a 
single  figure?  What  other  character  is  frequently  used?  Why  is  it 
sailed  naught?  How  many  are  the  Arabic  characters?  What  are 
lumber-  expressing  single  tilings  called  ? 


IT  4.  NOTATION  AND  NUMERATION.  11 

There  may  be  one,  two,  ,.3 

or  more  tens,  just  as  there  || 

are  one,  two,  or  more  units,     One  ten  is  .     .     .10  ten. 
or  single  things ;  it  takes     Two  tens  are   .     .    20  twenty, 
ten  cents  to  make  one  ten-     Three  tens  "    •     .30  thirty, 
cent  piece  ;  just  so  it  takes     Four  tens    "    .     .40  forty, 
ten  single  things  to  make     Five  tens     "    .     .50  fifty, 
one  ten.    All  figures  in  the     Six  tens       "    .     ,60  sixty. 
second  place  express  units     Seven  tens  "    .     .70  seventy, 
of  the  2d   order,  that  is,     Eight  tens  "    .     .80  eighty, 
units  of  tens.  Nine  tens    "    .     .90  ninety. 

One  ten  and  one  unit,  11,  One  ten,  one  unit,  11  e.  even, 
are  called  eleven ;  one  ten  One  ten,  two  units,  12  twelve, 
and  two  units,  12,  twelve, 

&c.  '  In  this  way  the  units  **<**•.  Tw™?>  thirty  &c  are 
of  the  1st  order  are  united  C&°f  actl°ns  f°r  tw°  tens' three  tens' 
with  the  tens,  that  is,  with 

the  units  of  the  2d  order,  to  form  the  numbers  from  10  to  20, 
from  20  to  30,  to  40,  and  so  on  to  99,  which  is  the  largest 
number  that  can  be  represented  by  two  figures. 

The  weeks  in  a  year  are  5  tens  and  2  units,  (5  of  the  sec 
ond  order  and  2  of  the  first  order  now  described,)  and  are 
expressed  thus,  52,  (fifty-two.)  In  the  same  manner  express 
on  your  slate,  or  on  the  blackboard,  the  two  orders  united,  so 
as  to  form  all  the  numbers  from  10  to  99. 

IT  4.    Ten  tens  are  called  one  hundred,  which  forms  a  unit 

of  a  still  higher,  or  3d  order,  and  is  ex-  ^»-» 

pressed   by  writing   two    ciphers  at  the  B£» 

right  hand  of  the  unit  1,     ...    thus,  100  one  hundred. 

Note.    When  there  are  no  units  or  tens,  we  200  two  hundred, 

write  ciphers  in  their  places,  which  denote  the  300  three  hundred, 
absence  of  a  thing,  (*f[  2.)  &c. 

Questions.  —  IT  3.  How  is  ten  represented?  What  is  it  considered 
as  forming?  Consisting  of  what?  What  place  does  the  cipher  fill? 
The  one?  Where  is  unit's  place,  and  where  ten's  place,  counting  from 
the  right  ?  How  much  larger  is  the  value  of  a  figure  in  the  place  of 
tens  than  in  the  place  of  units  ?  In  which  place  does  it  retain  its  sim 
pie  value  ?  In  ten's  place,  what  is  its  value  called?  What  is  1  ten  and 
1  unit  called  ?  1  ten  and  2  units  ?  How  are  the  numbers  from  10  to  99 
expressed?  Of  what  is  the  number  forty  made  up?  Ans.  4  tens  and 
no  units.  Sixty  ?  What  do  you  unite,  to  form  the  number  twenty, 
three?  thirty-seven?  seventy-five?  &c.  Of  what  are  twenty,  thirty, 
&c.,  contractions  ?  What  is  the  largest,  and  what  the  least,  number  you 
can  express  by  one  figure?  by  two  figures? 


12  NOTATION  AND  NUMERATION.  1T  5, 6. 

Three  hundred  sixty-five,  the  days  in  a  year,  are  expressed 
thus,  365 ;  3  being  in  the  place  of  hundreds,  6  in  the  place 
of  tens,  and  5  in  the  place  of  units. 

After  the  same  manner,  the  pupil  may  be  required  to  unite 
the  three  orders,  and  express  any  number  from  99  to  999. 

^F  O.  We  have  seen  that  figures  have  two  values,  viz.. 
simple  and  local. 

The  simple  value  of  a  figure  is  its  value  when  standing 
alone  ;  thus,  the  simple  value  of  7  is  seven. 

The  local  value  of  a  figure  is  its  value  according  to  its  dis 
tance  from  the  place  of  units  ;  thus,  the  local  value  of  7,  in  the 
number  75,  is  7  tens,  or  seventy,  while  its  simple  value  is 
seven ;  in  the  number  756,  its  local  value  is  seven  hundred. 

Note.  From  the  fact  that  10  is  1  more  than  9,  it  follows,  as  may 
be  found  by  trial,  that  the  local  value  of  every  figure  at  the  left  of 
units,  except  9,  exceeds  a  certain  number  of  nines  by  the  simple  value 
of  the  figure.  Take  the  number  623 ;  2  (tens)  is  2  more  than  2 
nines,  and  6,  (hundreds,)  6  more  than  a  certain  number  of  nines.  On 
this  principle  is  founded  a  method  of  proof  in  the  subsequent  rules,  by 
casting  out  the  nines. 

5F  6.  Ten  hundred  make  one  thousand,  which  is  called  a 
unit  of  the  next  higher,  or  4th  order,  consisting  of  thousands, 
and  is  expressed  by  writing  three  ciphers  at  the  right  hand  of 
the  unit  1,  giving  it  a  new  local  value  ;  thus,  1000,  one  thou 
sand.  ' 

To  thousands  succeed  tens  and  hundreds  of  thousands, 
forming  units  of  the  5th  and  6th  orders. 

Questions.  —  f  4.  What  are  10  tens  called  ?  What  do  they  form  ? 
How  many  places  are  required  to  express  hundreds  ?  How  much  does 
I  cipher,  placed  at  the  right  hand  of  1,  increase  it?  2  ciphers?  How 
do  you  express  two  hundred  ?  &c.  What  are  4  hundreds,  9  tens,  and  5 
units  called  ?  How  is  one  hundred  ninety-three  expressed  ?  What 
place  does  the  3  occupy  ?  the  9  ?  the  1  ?  How  do  you  express  the  ab 
sence  of  an  order?  How  is  the  number  of  days  in  a  year  expressed  ? 

If  5.  How  many  values  have  figures?  What  are  they?  What  is 
the  simple  value  ?  local  value  ?  What  is  the  value  of  5  in  59  ?  Is  it 
its  simple,  or  a  local  value  ?  Is  the  value  of  8,  in  874,  simple  or  local  ? 
of  the  7?  of  the  4? 

U  6.  How  do  you  express  one  thousand?  seven  thousand?  A  thou 
sand  is  a  unit  of  what  order  ?  How  many  thousands  are  30  hundreds  ? 
What  after  thousands  and  of  what  order  ?  The  6th  order  is  what  ?  In 
writing  nine  hundrel  and  two  thousand  and  nine,  where  do  you  place 
ciphers  ?  Why  ? 


IF  7, 8.  NOTATION  AND  NUMERATION.  13 

^F  7.    In  this  table  of  the  six  orders  now  _.  TABLE. 

described,  you  see  the  unit  1  moving  from  j 

right  to  left,  and  at  each  removal  forming  the  |  | 

unit  of  a  higher  order.     There  are  other  or-  jjj  f 

ders  yet  undescribed,  to  form  which  the  unit  -5  ^  |  | 

1  moves  onward  still  towards  the  left,  its  value  |  |    \  |    »   3 

being  increased  ten  times  by  each  removal.  a  £  &  »  &  5 

Note  1.    The  Ordinal  numbers,  1st,  2d,  3d,  &c.,  |   |   |   |   |   | 

may  be  called  indices  of  their  respective  orders.  g   -2   5   -a   g    « 

Afate  2.      Various   Readings.      In    the    number  ] 

546873,  the  left  hand  figure  5  expresses  5  units  of  j   Q 

the  6th  order,  or  it  may  be  rendered  in  the  next  100 

'ower  order  with  the  4,  and  together  they  may  be  i    n  n  n 
read  54  units  of  the  5th  order,  (ten  thousands,)  and 

connecting  with  the  6,  they  may  be  read,  546  units  10000 

of  the  4th  order,  or  546000.     Hence,  units  of  any  100000 

higher  order  may  be  rendered  in  units  of  any  lower  ,    ,    ,    ,    ,    , 

order.  999999 

To.  hundreds  of  thousands  succeed  units,  tens,  and  hun 
dreds  of  millions. 

^F  8.  To  millions  succeed  billions,  trillions,  quadrillions, 
quintillions,  sextillions,  septillions,  octillions,  nonillions,  decil- 
lions,  undecillions,  duodecillions,  tredecillions,  &c.,to  each  of 
which,  as  to  units,  to  thousands,  and  to  millions,  are  assigned 
three  places,  viz.,  units,  tens,  hundreds,  as  in  the  following  ex 
amples  : 

Questions.  —  ^[7.  How  is  the  unit  1  of  the  1st  order  made  a  unit 
of  the  2d  order?  of  the  3d  order,  &c.,  to  the  6th  order  ?  What  may  the 
ordinal  numbers,  1st,  2d,  3d,  &c.,  be  called  ?  7  units  of  the  6th  order 
are  how  many  units  of  the  4th  order  ?  The  teacher  mill  multiply  such 
questions.  What  is  the  least,  and  what  the  largest,  number  which  can 
be  expressed  by  2  places?  3  places?  fcc.  What  after  hundreds  of  thou 
sands?  Of  what  order  will  millions  be?  tens  of  millions?  hundreds  of 
millions? 

^[  8.  What  after  millions  ?  How  many  places  are  allotted  to  bil 
lions  ?  to  trillions  ?  &c.  Give  the  names  of  the  orders  after  trillions. 
In  reading  large  numbers,  \\hat  is  frequently  done?  Why?  The  1st 
period  at  the  right  is  the  period  of  what?  the  2d?  the  3d?  the  4th? 


14  NOTATION  AND   NUMERATION.  IT  9. 


2  .-§  H  g  .-§  I?  g  .-§  "c  S  .-§  1=  S  .5  15  c  .*  la  c  ••§ 

<D  'S    3  o>  c    3  <u  c    3  <u  c    PCS    s  <D  c    ^tyfi 

HPWHPKHPMEH^WH^KEH^ffiH^ 
3    082    715    203    174    592    837    463    512 

3,  0  8  2,  7  1  5,  2  0  3,  1  7  4,  5  9  2,  8  3  7,  4  6  3,  5  1  2 


3     3      1      I      j 

'S  S  M  .g  .g  -5 

To  facilitate  the  reading  of  large  numbers,  we  may  point 
them  off  into  periods  of  three  figures  each,  as  in  the  2d  exam 
ple.  The  names  and  the  order  of  the  periods  being  known, 
this  division  enables  us  to  read  numbers  consisting  of  many 
figures  as  easily  as  we  can  rea^l  those  of  only  three  figures. 
Thus,  in  looking  at  the  above  examples,  we  find  the  first  pe 
riod  at  the  left  hand  to  contain  one  figure  only,  viz.,  3.  By 
looking  under  it,  we  see  that  it  stands  in  the  9th  period  from 
units,  which  is  the  period  of  septillions ;  therefore  we  read  it 
3  septillions,  and  so  on,  82  sextillions,  715  quintillions,  203 
quadrillions,  174  trillions,  592  billions,  837  millions,  463  thou 
sands,  512. 

^F  9.  From  the  foregoing  we  deduce  the  following  princi 
ples  :  — 

Numbers  increase  from  right  to  left,  and  decrease  from  left 
to  right,  in  a  ten-fold  proportion ;  and  it  is 

A  FUNDAMENTAL  LAW  OF  THE  ARABIC  NOTATION  ;  that, 

Questions.  —  H  9»  How  do  numbers  increase  ?  how  decrease,  and 
in  what  proportion  ?  To  what  is  1  ten  equal  ?  1  hundred?  1  thousand? 
&c.  To  what  are  10  units  equal?  10  hundreds  ?  &c.  What  is  a  fun 
damental  law  of  the  Arabic  notation  ?  What  is  notation  ?  numera 
tion  ?  How  do  you  write  numbers  ?  read  numbers  ?  If  you  were  to 
write  a  number  containing  units,  tens,  hur  ireds,  and  millions,  but  no 
thousands,  how  would  you  express  it  ? 


1F  10.  NOTATION  AND  NUMERATION.  15 

I.  Removing  any  figure  one  place  towards  the  left,  in 
creases  its  value  ten  times,  and 

II.  Removing  any  figure  one  place  towards  the  right,  de 
creases  its  value  ten  times. 

The  expressing  of  numbers  as  now  shown  is  called  Nota 
tion.  The  reading  of  any  number  set  down  in  figures  is 
called  Numeration. 

To  write  numbers.  —  Begin  at  the  left  hand,  and  write  in 
their  respective  places  the  units  of  each  order  mentioned  in 
the  number.  If  any  of  the  intermediate  orders  of  units  be 
omitted  in  the  number  mentioned,  supply  their  respective 
places  with  ciphers. 

To  read  numbers.  —  Point  them  off  into  periods  of  three 
figures  each,  beginning  at  the  right  hand ;  then,  beginning  at 
the  left  hand,  read  each  period  separately. 

Let  the  pupil  write  down  and  read  the  following  numbers : 

Two  million,  eighty,  thousand,  seven  hundred  and  five. 
One  hundred  million,  one  hundred  thousand  and  one. 
Fifty-two  million,  sixty  thousand,  seven  hundred  and  three. 
One  hundred  thirty-two  billion,  twenty-seven  million. 
Five  trillion,  sixty  billion,  twenty-seven  million. 
Seven  hundred  trillion,  eighty-six  billion,  and  nine. 
Twenty-six  thousand,  five  hundred  and  fifty  men. 
Two  million,  four  hundred  thousand  dollars. 
Ninety-four  billion,  eighty  thousand  minutes. 
Sixty  trillion,  nine  hundred  thousand  miles. 

Eighty-four  quintillion,  seven  quadrillion,  one  hundred  million 
grains  of -sand. 

5T  1O.    Numbers  are  employed  to  express  quantity. 

Quantity  is  anything  which  can  be  measured.  Thus, 
Time  is  quantity,  as  we  can  measure  a  portion  of  it  by  days, 
hours,  &c.  Distance  is  quantity,  as  it  can  be  measured  by 
miles,  rods,  &c. 

By  the  aid  of  numbers  quantities  may  either  be  added  to 
gether,  or  one  quantity  may  be  taken  from  another. 

Arithmetic  is  the  art  of  making  calculations  upon  quanti 
ties  by  means  of  numbers. 

Questions. — IT  1O.  Numbers  are  employed  to  express  what? 
What  is  quantity  ?  By  what  is  a  quantity  of  grain  measured  ?  a  quan 
tity  of  cloth-?  What  is  arithmetic  ?  What  is  an  abstract  number  ?  a 
denominate  number?  What  is  the  unit  of  a  number  1  What  is  the 
unit  value  of  8  bushels  ?  of  16  yards  ?  of  20  pounds  of  sugar  ?  of  3 
quarts  of  milk?  of  9  dozen  of  buttons?  of  18  tons  of  hay  ?  of  16  hogs 
heads  of  molasses  ? 


16  ADDITION  OF  SIMPLE   NUMBERS.  H  11. 

A  number  applied  to  no  kind  of  thing,  as  5,  10,  18,  36,  is 
called  an  abstract  number. 

A  number  applied  to  some  kind  of  thing,  as  7  horses,  25 
dollars,  250  men,  is  called  a  denominate  number. 

The  unit,  or  unit  value  of  a  number,  is  one  of  the  kind 
\vhich  the  number  expresses ;  thus,  the  unit  of  99  days  is  1 
day ;  the  unit  of  7  dollars  is  1  dollar ;  the  unit  of  15  acres  is 
1  acre.  In  like  manner  the  unit  of  9  tens  may  be  said  to  be 
1  ten ;  the  unit  of  8  hundred  to  be  1  hundred ;  the  unit  of  6 
thousand  to  be  1  thousand,  &c. 


ADDITION   OF   SIMPLE  NUMBERS. 

^Tll.  1.  James  had  5  peaches,  his  mother  gave  him  3 
more  ;  how  many  had  he  then  ?  Ans.  8. 

Why  ?     Ans.  Because  5  and  3  are  8. 

2.  Henry,  in  one  week,  got  17  merit  marks  for  perfect  les 
sons,  and  6  for  good  behavior ;  how  many  merit  marks  did 
he  get  ?  Ans.  .    Why  ? 

3.  Peter  bought  a  wagon  for  36  cents,  and  sold  it  so  as  to 
gain  9  cents ;  how  many  cents  did  he  get  for  it  ? 

4.  Frank  gave  15  walnuts  to  one  boy,  8  to  another,  and 
had  7  left;  how  many  walnuts  had  he  at  first  ? 

5.  A  man  bought  a  chaise  for  54  dollars ;  he  expended  8 
dollars  in  repairs,  and  then  sold  it  so  as  to  gain  5  dollars; 
how  many  dollars  did  he  get  for  the  chaise  ? 

The  putting  together  of  two  or  more  numbers,'  (as  in  the 
foregoing  examples,)  so  as  to  make  one  whole  number,  is 
called  Addition,  and  the  whole  number  is  called  the  Sum,  or 
Amount. 

6.  One  man  owes  me  5  dollars,  another  owes  me  6  dol 
lars,  another  8  dollars,  another  14  dollars,  and  another  3  dol 
lars  ;  what  is  the  sum  due  to  me  ? 

7.  What  is  the  amount  of  4,  3,  7,  2,  8,  and  9  dollars  ? 

8.  In  a  certain  school,  9  study  grammar,  15  study  arith 
metic,  20  attend  to  writing,  and  12  study  geography ;  what 
is  the  whole  number  of  scholars  ? 


Questions. — II  11.  What  is  addition?  What  is  the  answer,  01 
number  sought,  called?  What  is  the  sign  of  addition?  What  does  it 
show?  How  is  it  sometimes  read?  Whence  the  word  plus,  and  what 
'.s  its  signification?  What  is  the  sign  of  equality,  and  what  does  it 


Til. 


ADDITION  OF  SIMPLE  NUMBERS. 


17 


SIGNS.  —  A  cross,  -)-,  one  line  horizontal  and  the  other  per 
pendicular,  is  the  sign  of  Addition.  It  shows  that  numbers 
vvith  this  sign  between  them  are  to  be  added  together ;  thus, 
4_^_7_j_14_(_16  denote  that  4,  7,  14,  and  16  are  to  be 
added  together.  It  is  sometimes  read  plus,  which  is  a  Latin 
word  signifying  more. 

Two  parallel,  horizontal  lines,  =,  are  the  sign  of  Equality. 
It  signifies  that  the  number  before  it  is  equal  to  the  number 
after  it ;  thus,  5-}- 3  ==8  is  read  5  and  3  are  8;  or,  5  plus  3 
are  equal  to  8. 

In  this  manner  let  the  pupil  be  instructed  to  commit  the 
following 

ADDITION  TABLE. 


2- 

-0=   2 

3- 

rO=     3 

4- 

|-0=   4 

5- 

hO=    5 

2- 

-1  =   3 

3- 

-1=    4 

4- 

-1=    5 

5-^ 

-1=   6 

2- 

-2=    4 

3- 

r2=   5 

4- 

-2=    6 

5- 

-2=    7 

2- 

-3=    5 

3- 

-3=   6 

4- 

-3=    7 

5- 

-3=   8 

2- 

-4=    6 

3- 

-4=    7 

4- 

-4=    8 

5- 

-4=    9 

2- 

-5  =    7 

3- 

-5=    8 

4- 

—  fj  =  y 

5- 

r5=10 

2- 

-6=    8 

3- 

-6=    9 

4- 

-6  =  10 

5- 

-6  =  11 

2- 

-7=    9 

3- 

-7=10 

4- 

-7  =  11 

5- 

-7  =  12 

2- 

-8=10 

3- 

-8=11 

4- 

-8=12 

5- 

-8=13 

2- 

-9  =  11 

3- 

_  9  .  —  ,12 

4- 

-9=13 

5- 

-9=14 

6- 

-0=    6 

7- 

[-0=    7 

8- 

-0=   8 

9-4-0=    9 

6- 

-  1  =    7 

7- 

-1=    8 

8- 

-1=   9 

9- 

-1  =  10 

6- 

-2=   8 

7- 

-2=    9 

8- 

-2=10 

9- 

-2=11 

6- 

-3=    9 

7- 

-3=10 

8- 

-3=11 

9- 

-3=12 

6- 

-4=10 

7- 

-4=11 

8- 

-4=12 

9- 

-4=13 

6- 

-5=11 

7- 

_5  —  12 

8- 

-5=13 

9- 

-5=14 

6- 

-6=12 

7- 

-  6  =  13 

8- 

-6=14 

9- 

-6=15 

6- 

-  7  =  13 

7- 

-7=14 

8- 

-7=15 

9- 

-7=16 

6- 

-  8  =  14 

7- 

-8=15 

8- 

-8=16 

9- 

-8=17 

6- 

-9=15 

7- 

-9=16 

8- 

-9=17 

9- 

-9  =  18 

•  9  =  how  many  ? 

•  7  =  how  many  ? 

•  3  -f-  2  =  how  many  ? 

•  4  -  -  5  =  how  many  ? 

•  0  -  -  4  -j-  6  =  how  many  ? 

•  1  -  -  0  -j-  8  =  how  many  ? 

•  0  -I-  9  4-  5  =  how  many  ? 

2* 


18 


ADDITION   OF   SIMPLE   NUMBERS. 


1F12 


9_L_2-f-6-f-4-|-5==  how  many  ? 
l__3_-5--7--S  =  how  many  ? 
1  __9__3__4_-/>-|-  6  =  how  many ? 
8__9__0--2--4--  5  =  how  many? 
64-2--5-j-0-|-S-|-3  =  how  many ? 


IF  1^1.  When  the  numbers  to  be  added  are  small,  the  ad 
dition  is  readily  performed  in  the  mind,  and  this  is  called 
mental  arithmetic  ;  but  it  will  frequently  be  more  convenient, 
and  even  necessary,  when  the  numbers  are  large,  to  write 
them  down  before  adding  them,  and  this  is  called  written 
arithmetic. 

1.  Harry  had  43  cents,  his  father  gave  him  25  cents  more ; 
how  many  cents  had  he  then  ? 

SOLUTION. — One  of  these  numbers  contains  4  tens  and  3  units. 
The  other  number  contains  2  tens  and  5  units.     To  unite  these  two 
numbers  together  into  one,  write  them  down  one  un- 
43  cents.      der  the  other,  placing  the  units  of  one  number  directly 
25  cents.      under  units  of  the  other,  and  the  tens  of  one  number 
directly  under  tens  of  the  other,  and  draw  a  line  un 
derneath. 


43  cents. 
25  cents. 


43  cents. 
25  cents. 

Ans.  68  cents. 


Beginning  at  the  column  of  units,  we  add  each 
column  separately ;  thus,  5  units  and  3  units  are  8 
units,  which  we  set  down  in  units'  place. 

We  then  proceed  to  the  column  of  tens,  and  say, 
2  tens  and  4  tens  are  6  tens,  or  60,  which  we  set 
down  directly  under  the  column  in  tens'  place,  and 
the  work  is  done. 


It  now  appears  that  Harry's  whole  number  of  cents  is  6 
tens  and  8  units,  or  68  cents  ;  that  is,  43  -|-  25  =  68. 

Units  are  written  under  units,  tens  under  tens,  &c. ;  be 
cause  none  but  figures  of  the  same  unit  value  can  be  added  to 
each  other ;  for  5  units  and  3  tens  will  make  neither  8  tens 
nor  8  units,  just  as  5  cows  and  3  sheep  will  make  neither  8 
cows  nor  8  sheep. 

Questions. —  ^[  12.  What  distinction  do  you  make  between  men 
tal  and  tv ritten  arithmetic ?  How  do  you  write  numbers  for  addition? 
Where  do  you  begin  to  add  ?  and  where  do  you  set  the  amount  ?  Ho\v 
do  you  proceed?  Why  do  you  write  units  under  units,  tens  under  tens, 
Ace.?  * 


f  13.  ADDITION   OF  SIMPLE  NUMBERS.  19 

2.  A  farmer  bought  a  chaise  for  210  dollars,  a  horse  foi 
70  dollars,  and  a  saddle  for  9  dollars ;  what  was  the  whole 
amount  ? 

Write  the  numbers  as  before  directed,  with  units  under 
units,  tens  under  tens,  &c. 

OPERATION. 

Chaise.  210  dollars. 

Hnrw       70  dollar*  Add  as  before-     The  units  Wl11  be  9»  the 

o>5j        o  jjj  tens  8,  and  the  hundreds  2  ;  that  is,  210 -|- 

Saddle,      9  dollars.  70  _l_  9  =  289. 

Answer,  289  dollars. 

After  the  same  manner  are  performed  the  following  exam 
ples,  in  which  the  amount  of  no  column  exceeds  nine. 

3.  A  man  had  15  sheep  in  one  pasture,  20  in  another  pas 
ture,  and   143  in  another ;    how  many  sheep  had  he  in  the 
three  pastures  ?  15  -J-  20  -j-  143  =  how  many  ? 

4.  A  man  has  three  farms,  one  containing  500  acres,  an 
other  213  acres,  and  another  76  acres ;  how  many  acres  in 
the  three  farms  ?     500  +  213  +  76  =  how  many  ? 

5.  Bought  a  farm  for  2316  dollars,  and  afterwards  sold  it 
so  as  to  gain  550  dollars ;  what  did  I  sell  the  farm  for  ?    2316 
-{-  550  =  how  many  ? 

6.  A  chair-maker  sold,  in  one  week,  30  Windsor  chairs,  36 
cottage,  102  fancy,  and  21  Grecian  chairs  ;    how  many  chairs 
did  he  sell  ?     30  +  36  +  102  +  21  =  how  many  ? 

7.  A  farmer,  after  selling  500  bushels  of  wheat  to  a  com 
mission  merchant,  320  to  a  miller,  and  sowing  117  bushels, 
found  he  had  62  bushels  left;  how  many  bushels  had  he  at 
first  ?     500  +  320  -f- 1 17  +  62  =  how  many  ? 

8.  A  dairyman  carried  to  market  at  one  time  231  pounds 
of  butter,  at  another  time  124,  at  another  302,  at  another  20, 
and  at  another  12 ;  how  many  pounds  did  he  carry  in  all  ? 

Ans.  689  pounds. 

9.  A  box   contains    115  arithmetics,  240  grammars,  311 
geographies,  200  reading  books,  and  133  spelling  books ;  how 
many  books  are  there  in  the  box?  Ans.  999. 

IT  13.  Hitherto  the  amount  of  any  one  column,  when 
added  up,  has  not  exceeded  9,  ai  d  consequently  has  been  ex 
pressed  by  a  single  figure.  But  it  will  frequently  happen 
that  the  amount  of  a  single  column  will  exceed  9,  requiring 
two  or  more  figures  to  express  it. 

1.    There  are  three  bags  of  money.     The  first  contains  876 


^0  ADDITION  OP  SIMPLE  NUMBERS.  IT  13. 

dollars,  the  second  653  dollars,  the  third  426  dollars  ;  what  is 
the  amount  contained  in  all  the  bags  ? 

OPERATION.  SOLUTION.  — Writing  the  numbers  as 

First  bag,    876  dollars,      already  described,  we  add  the  units,  and 
c          ?  u       gco        a  find  them  to  be  15,  equal  to  5  units,  winch 

T7  '  //    "       /L9fi         '  we  wr*te  *n  umts'  place,  adding  the  1 

ten  with  the  tens  ;  which  being   added 
together  are  15  tens,  equal  to  5  tens,  to 
1955        "  be  written  in  tens'  place,  and  1  hundred, 

to  be  added  to  the  hundreds.  The  hun 
dreds  being  added  are  19,  equal  to  9  hundreds,  to  be  written  in  hun 
dreds'  place,  and  1  thousand,  to  be  written  in  thousands'  place. 

Ans.   1955  dollars. 

PROOF. — We  may  reverse  the  order,  and,  beginning  at  the  top,  add 
the  figures  downwards.  If  the  two  results  are  alike,  the  work  may 
be  supposed  to  be  right,  for  it  is  not  likely  that  the  same  mistake  will 
be  made  twice,  when  the  figures  are  added  in  a  different  order. 

NOTE.  —  Proof  by  the  excess  of  nines.  If  the  work  be  right,  there 
will  be  just  as  many  of  any  small  number,  as  9,  with  the  same  re 
mainder,  in  the  amount,  as  in  the  several  numbers  taken  together. 
Hence, 

In  the  upper  number,  8  (hundreds)  is  8  more  than  a 
J~'     certain  number  of  nines,    (*|[  5)   7   (tens)  is  7  more. 


OPERATION. 


876 
653 

426 


1955 


3 


,.  Adding  the  8  and  7,  and  the  6  units  together,  the  sum 
is  21  =  2  nines  and  3  remainder,  which  we  set  down  at 
the  right  hand,  as  the  excess  of  nines  in  this  number. 
In  the  same  manner,  5  is  found  to  be  the  excess  of  nines 
in  the  second  number,  and  3  in  the  third  number.  These 
several  excesses  being  added  together,  make  1  nine  and  an  excess  of 
2,  which  is  the  same  as  the  excess  of  nines  in  the  general  amount, 
found  in  the  same  manner.  This  method  will  detect  every  mistake, 
except  it  be  9,  or  an  exact  number  of  nines. 

To  find  what  will  be  the  excess  after  casting  the  nines  out  of  any 
number,  begin  at  the  left  hand,  and  add  together  the  figures  ivhich 
express  the  number;  thus,  to  cast  the  nines  out  of  892,  we  say  8 
(passing  over  9)  -f-2  (dropping  9  from  the  sum)  =  1. 

From  the  examples  and  illustrations  now  given,  we  derive 
the  following 

RULE. 

I.   Write  the  numbers  to  be  added,  one  under  another,  plac- 


Qtiestions.  — TI  13.  If  the  amount  of  the  column  does  not  exceed 
9,  what  do  you  do?  What  wheT.  it  exceeds  9 ?  How  do  you  add  each 
column?  What  do  you  do  witn  the  amount  of  the  left  column?  For 
what  number  do  you  carry  ?  If  the  amount  of  a  column  be  36,  what 
would  you  set  down,  and  how  many  would  you  carry  ?  On  what  prin 
ciple  do  you  do  this?  How  is  addition  proved ?  Why?  Repeat  the 
rule  for  addition. 


IF  13.  ADDITION  OF  SIMPLE  NUMBERS.  21 

ing  units  under  units,  tens  under  tens,  &c.,  and  draw  a  line 
underneath. 

II.  Begin  at  the  unit  column,  and  add  together  all  the  fig 
ures  contained  in  it;  if  the  amount  does  not  exceed  9,  write 
it  under  the  column ;  but  if  it  exceed  9,  write  the  units  in 
units'  place,  and  carry  the  tens  to  the  column  of  tens. 

III.  Add  each  succeeding  column  in  the  same  manner,  and 
set  down  the  whole  amount  of  the  last  column. 

EXAMPLES    FOR    PRACTICE. 

J5l  — 

2863705421061  4367583021463 

3107429315638  1752349713620 

6253034792  6081275306217 

247135  5652174630128 

8673  8703263472013 

4.  Add  together  587,  9658,  67,  431,  28670,  85,  100000, 
6300,  and  1.  Amount,  145799. 

5.  What  is  the  amount  of  8635,  7,  2194,  16,  7421.  93, 
5063,  135,  2196,  89,  and  1225?  Ans.  27074. 

6.  A  man  being  asked  his  age,  answered  that  he  left  Eng 
land  when  he  was  12  years  old,  and  that  he  had  afterwards 
spent  5  years  in  Holland,  17   years  in  Germany,  9  years  in 
France,  whence  he  sailed  for  the  United  States  in  the  year 
1827,  where  he  had  lived  22  years ;  what  was  his  age  ? 

Ans.  65  years. 

7.  A  company' contract  to  build  six  warehouses;  for  the 
first  *  they  receive  36850  dolls. ;  for  the  second,  43476  dolls. ; 
for  the  third,  18964  dolls. ;  for  the  fourth,  62840  dolls. ;  for  the 
fifth,  71500  dolls. ;  for  the  sixth,  as  much  as  for  the  first  three ; 
to  what  do  these  contracts  amount  ?     Aris.  332920  dollars. 

8.  James  had  7  marbles,  Peter  had  4  marbles  more  than 
James,  and  John  had  5  more  than  Peter  ;  how  many  marbles 
in  all  ?  Ans.  34. 

.  9.  There  are  seven  men  ;  the  first  man  is  worth  67850  dol 
lars  ;  the  second  man  is  worth  2500  dolls,  more  than  the  first 
man  ;  the  third,  3168  dolls,  more  than  the  second  ;  the  fourth, 
16973  dolls,  more  than  the  third ;  the  fifth,  40600  dolls,  more 
than  the  fourth ;  the  sixth,  19888  dolls,  more  than  the  fifth  ; 
and  the  seventh,  49676  dolls,  more  than  the  sixth  ;  how  many 
dollars  are  they  all  worth  ?  Ans.  784934  dollars. 

10.    What  is  the  interval  in  years  between  a  transaction 


22  ADDITION  OF  SIMPLE  NUMBERS.  IT  14. 

that  happened  275  years  ago,  and  one  that  will  happen  125 
years  hence  ?  Ans.  400  years. 

11.  What  is  the  amount  of  46723,  6742,  and  986  dollars  ? 

12.  A  man  has  three  orchards ;  in  the  first  there  are  140 
trees  that  bear  apples,  and  64  trees  that  hear  cherries ;  in  the 
second,  234  trees  bear  apples,  and  73  bear  cherries ;  in  the 
third,  47  trees  bear  plums,  36  bear  pears,  and  25  bear  cher 
ries  ;  how  many  trees  in  all  the  orchards,  and  how  many  of 
each  kind  ? 

Am.  619  trees;  374  bear  apples ;  162  cherries ;  47  plums; 
and  36  pears.  . 

13.  A  gentleman  purchased  a  farm  for  7854  dollars  ;  he 
paid  194  dollars  for  having  it  drained  and   fenced,  and  300 
dollars  for  having  a  bam  built  upon  it ;  how  much  did  it  cost 
him,  and  for  how  much  must  he  sell  it,  to  gain  273  dollars  ? 

It  cost  him  8348  dollars. 

He  must  sell  it  for  8621  dollars. 


IT  14.    Review  of  Numeration  and  Addition. 

Questions.  —  What  are  numbers?  What  are  the  methods  of 
expressing  numbers?  What  is  numeration?  notation?  fundamental 
law  in  the  Arabic  notation?  How  does  the  Arabic  differ  from  the  Ro 
man  method?  What  is  understood  by  units  of  different  orders  ?  What 
is  quantity?  Arithmetic?  What  is  understood  by  the  simple  value  of 
figures?  the  local  value  ?  the  unit  value  of  a  number?  Explain  the 
difference  between  an  abstract  and  a  denominate  number.  What  is 
addition?  the  rule?  proof?  For  what  number  do  you  carry,  and  why? 

EXERCISES. 

1.  Washington  was  born  in  the  year  of  our  Lord  1732  ; 
he  was  67  years  old  when  he  died ;  in  what  year  did  he  die  ? 

Ans.   1799. 

2.  The  invasion  of  Greece  by  Xerxes  took  place  481  years 
before  Christ ;  how  long  ago  is  that  this  current  year  ? 

3.  There  are  two  numbers  ;  the  less  is  8671,  the  difference 
between  the  numbers  is  597 ;  what  is  the  greater  number  ? 

Ans.  9268. 

4.  A  man  borrowed  a  sum  of  money,  and  paid  in  part  684 
dollars ;  the  sum  left  unpaid  was  876  dollars ;  what  was  the 
sum  borrowed? 

5.  There  are  four  numbers  ;  the  first  317,  the  second  812 
the  third  1350,  and  the  fourth  as  much  as  the  other  three ; 
what  is  the  sum  of  them  all  ?  Ans.  4958. 

6.  A  gentleman  left  his  daughter  16  thousand  16  hundred 


IT  15.  SUBTRACTION  OF  SIMPLE  NUMBERS.  23 

and  16  dollars;  he  left  his  son  1800  more  than  his  daughter; 
what  was  his  son's  portion,  and  what  was  the  amount  of  the 
whole  estate  ?  A  (  Son's  portion,  19416. 

Am'    I  Whole  estate,  37032. 

7.  A  man,  at  his  death,  left  his  estate  to  his  four  children, 
who,  after  paying  debts  to  the  amount  of  1476  dollars,  re 
ceived  4768  dollars  each  ;  how  much  was  the  whole  estate  ? 

A?is.  20548. 

8.  A  man  bought  four  hogs,  each  weighing  375  pounds ; 
how  much  did  they  all  weigh  ?  A?is.   1500. 

9.  The   fore   quarters  of  an  ox  weigh  one  hundred  and 
eight  pounds  each,  the  hind  quarters  weigh  one  hundred  and 
twenty-four  pounds   each,  the  hide   seventy-six  pounds,  and 
the  tallow  sixty  pounds  ;  what  is  the  whole  weight  of  the  ox  ? 

Ans.  600, 

10.  The  imports  into  the  several  States  in  1842  were  as 
follows  :  Me.  606864  dollars,  N.  H.  60481,  Vt.  209868,  Mass. 
17986433,  R.  I.  323692,  Ct.  335707,  N.  Y.  57875604,  N.  J. 
145,  Pa.  7385858,  Del.  3557,  Md.  4417078.  D.  C.  29056,  Va. 
316705,   N.  C.   187404,    S.  C.   1359465,   Ga.   341764,    Al. 
363871.    La.  8033590,    O.  13051,    Ky.  17306,    Term.  5687, 
Mich.  80784,  Mo.  31137,  Fa.  176980  dollars ;    what  was  the 
entire  amount  ?  Ans.  100162087. 


SUBTRACTION   OF   SIMPLE   NUMBERS. 

5T  15.     1.    Charles,  having  18  cents,  bought  a  book,  for 
which  he  gave  6  cents  ;  how  many  cents  had  he  left? 

2.  John  had  12  apples  ;  he  gave  5  of  them  to  his  brother; 
how  many  had  he  left  ? 

3.  Peter  played  at  marbles ;  he  had  23  when  he  began, 
but  when  he  had  done  he  had  only  12;  how  many  did  he 
lose  ? 

4.  A  man  bought  a  cow  for  17  dollars,  and  sold  her  again 
for  22  dollars  ;  how  many  dollars  did  he  gain  ? 

5.  Charles  i?  9  years  old,  and  Andrew  is  13;  what  is  the 
difference  in  their  ages  ? 

6.  A  man  borrowed  50  dollars,  and  paid  all  but  18;  how 
many  dollars  did  he  pay  ?  that  is,  take  18  from  50,  and  how 
many  would  there  be  left  ? 

The  taking  of  a  less  number  from  a  greater  (as  in  the  fore 
going  examples)  is  called  Subtraction.     The  greater  number 


24 


SUBTRACTION  OF  SIMPLE  NUMBERS. 


If  16. 


is  called  the  Minuend,  the  less  number  the  Subtrahend,  and 
what  is  left  after  subtraction  is  called  the  Difference,  or  Hi 
mainder. 

7.  If  the  minuend  be  8,  and  the  subtrahend  3,  what  is  the 
difference  or  remainder  ?  Ans.  5. 

8.  If  the  subtrahend  be  4,  and  the  minuend  16,  what  is  the 
remainder  ? 

SIGN.  —  A  short  horizontal  line,  —  ,  is  the  sign  of  subtrac 
tion.  It  is  usually  read  minus,  which  is  a  Latin  word  signi 
fying  less.  It  shows  that  the  number  after  it  is  to  be  taken 
from  the  number  before  it.  Thus,  S  —  3  =  5  is  read  8 
minus  or  less  3  is  equal  to  5  ;  or,  3  from  8  leaves  5.  The 
latter  expression  is  to  be  used  by  the  pupil  in  committing  the 


following  • 


2  —  2  =  0 

3  —  2=1 

4  —  2  =  2  8  —  3  =  5 

5  —  2  =  3  9  —  3  =  6 

6  —  2  =  4  I  10  —  3  =  7 

7  —  2  =  5  4_4==o 


SUBTRACTION    TABLE. 


6  —  3  =  3 


8  —  2  =  6 

9  —  2  =  7 
10  —  2  =  8 


5  —  4=1 

6  —  4  =  2 


3  —  3  =  0        8  —  4  =  4 

4  —  3=1   !     9  —  4  =  5 

5  —  3  =  2  I   10  —  4  =  6   I   10  —  6  =  4 


5  —  5  =  0 
6  —  5=1 

7  —  5  =  2 

9_5  =  4 
10  —  5  =  5 

7  —  7  =  0 
8  —  7=1 
9  —  7  =  2 
10  —  7  =  & 

8  —  8  =  0 
9  —  8=1 
10  —  8  =  2 

6  —  6  =  0 
7  —  6=1 
8  —  6  =  2 
9  —  6  =  3 

9  —  9  =  0 
10  —  9  =  1 

7  —  3  =  how  many  ? 
8  —  5  =  how  many  ? 
9  —  \  =  how  many  ? 
12  -—  3  =  how  many  ? 
1£  —  4  =  how  many  ? 

18  —   7  =  how  many  ? 
28  —    7  =  how  many  ? 
22  —  13  =  how  many  ? 
33  —    5  =  how  many  ? 
41  —  15=  how  many  ? 

^1  11$.  When  the  numbers  are  small,  as  in  the  foregoing 
examples,  the  taking  of  a  less  number  from  a  greater  is 
readily  done  in  the  mind  ;  but  when  the  numbers  are  large, 

Questions.  —  *[[  15.  What  is  subtraction  ?  What  is  the  greater 
number  called?  the  less  number?  that  which  is  left  after  subtraction? 
What  is  the  sign  of  subtraction?  How  is  it  usually  read?  What  does 
minus  mean  {  What  does  the  sign  of  subtraction  show? 


U  16.  SUBTRACTION  OF  SIMPLE  NUMBERS*  25 

the  operation  is  most  easily  performed  part  at  a  time,  and 
therefore  it  is  necessary  to  write  the  numbers  down  beforo 
performing  the  operation. 

1.  A  farmer,  having  a  flock  of  237  sheep,  lost  114  of  them 
by  disease ;  how  many  had  he  left  ? 

Here  we  have  4  units  to  be  taken  from  7  units,  1  ten  to  be 
taken  from  3  tens,  and  1  hundred  to  be  taken  from  2  hundreds. 
It  will  therefore  be  most  convenient  to  write  the  less  number 
under  the  greater,  observing,  as  in  addition,  to  place  units 
under  units,  tens  under  tens,  &c.,  thus  : 

OPERATION.  SOLUTION.  —  We  begin   with   the 

From  237  the  minuend,  units  saying,  4  (units)  from?  (units,) 

rr  7      11/1  +  1,        i,      7      j  and  there  remain  3,  (units,)    which 

Take  1 14  the  subtrahend.  we  ^  down  directly  ^  ^  column 

~  .  in  units'  place.     Then  proceeding-  to 

123  the  remainder.      the  next  column,  we  say,  1  (ten)  from 

3,  (tens,)  and  there  remain  2,  (tens,) 

which  we  set  down  in  tens'  place.  Proceeding  to  the  next  column, 
we  say,  1  (hundred)  from  2,  (hundreds,)  and  there  remains  1,  (hun 
dred,)  which  we  set  down  in  hundreds'  clace,  and  the  work  is  done. 
It  now  appears  that  the  number  of  sheep  left  was  123  ;  that  is,  237 
—  114=123,  Ans. 

NOTE. — We  write  units  under  units,  tens  under  tens,  &c.,  that 
those  of  the  same  unit  value  may  be  subtracted  from  each  other ;  for 
we  can  no  more  take  3  tens  from  7  units  than  we  can  lake  3  cows 
from  7  sheep. 

Examples  in  which  each  figure  in  the  subtrahend  is  less  than 
the  figure  above  it. 

2.  There  are  two  farms;  one  is  valued  at  3750,  and  the 
other  at  1500  dollars ;  what  is  the  difference  in  the  value  of 
the  two  farms?      .  .  Ans.  2250. 

3.  A  mairs  property  is  worth  8560  dollars,  but  he  has  debts 
to  the  amount  of  3500  dollars  ;  what  will  remain  after  paying 
his  debts  ?  Ans.  5060. 

4.  From  746  subtract  435.  Rem.  311. 

5.  From  4983  subtract  2351.  Rem.  2632. 

6.  From  658495  subtract  336244.  Rem.  322251. 

7.  From  8764292  subtract  7653181.         Rem.  1111111. 


Questions.  —  ^f  16.  When  the  numbers  are  small,  hc\v  may  the 
operation  be  performed?  When  they  are  large,  what  is.  more  conve- 
t^ent  ?  How  are  the  two  numbers  to  be  written  ?  Where  do  you  begin 
***  subtraction  ? 

8 


26  SUBTRACTION  OF  SAMPLE  NUMBERS.  IT  17. 

f  17.     1.    James,  having  15  cents,  bought  a  pen-knife, 
for  which  he  gave  7  cents  ;  how  many  cents  had  he  left  ? 

OPERATION. 

15  cents.  A  difficulty  presents  itself  here;    for  we  cannot 

7  cents.  take  7  from  5  ;    but  we  can  take  7  from  15,  and 
_  there  will  remain  8. 

8  cents  left. 

2.   A  man  bought  a  horse  for  85  dollars,  and  a  cow  for  27 
dollars  ;  what  did  the  horse  cost  him  more  than  the  cow  ? 

OPERATION.         SOLUTION.  —  The  same  difficulty  presents  itself  hoie 

QQ  '  as  in  the  last  example,  that  is,  the  unit  figure  in  the 

27  subtrahend  is  greater  than  the  unit  figure  in  the  minu 

end.     To  obviate  this  difficulty,  we  may  take  1  (ten) 

from  the  8  (tens)  in  the  minuend,  which  will  leave 

7  (tens,)  and  add  it  to  the  5  units,  making  15  units,  (7  tens  -j-  15 

units  =  85,)  thus, 

TENS.    UNITS. 

7     15  We  now  take  7  units  from  15  unit^,  and  2  tens  from 

277  tens,  and  have  5  tens  and  8  units,  or  58  remainder  ; 
_         that  is  85  —  27  =  58  dollars  more  for  the  horse  than 

for  the  cow. 
O       o 

The  operation  may  be  shortened  as  follows  : 

OPERATION.  We  have  8  tens  and  5  units  in  the  minuend, 

IT          Qz  J  77  an(*  2  tens  an(l  7  units  in  the  subtrahend.     We 

j-iorse,  so  a     irs. 


>  ^  suppose  x  ten  taken  frora 

Low,  £1  the  8  tens,  which  would  leave  7  tens,  and  this 

1  ten  we  can  suppose  joined  to  the  5  units, 
Diff.  5S  "  making  15.  We  can  now  take  7  from  15,  as 

before,  and  there  will  remain  8,  which  we  set 

down.  The  taking  of  1  len  out  of  8  tens,  and  joining  it  with  the  5 
units,  is  called  borrowing  ten.  Proceeding  to  the  next  higher  order, 
or  tens,  we  must  consider  the  upper  figure,  8,  from  which  we  bor 
rowed,  .1  less,  calling  it  7  ;  then,  taking  2  (tens)  from  7,  (tens,)  there 
will  remain  5,  (tens,)  which  we  set  down,  making  the  difference  58 
dollars,  Ans. 


Questions.  —  ^[  17.  In  subtracting  7  from  15,  what  difficulty  pre 
sents  itself?  How  do  you  obviate  it  ?  In  taking  27  from  85,  instead  of 
taking  7  from  5  what  do  you  take  it  from?  Whence  the  15?  From 
what  do  you  subtract  the  2  tens  ?  Why  not  from  8  tens  instead  of  7 
tens?  What  is  this  operation  called?  ExpJain  how  the  operation  is 
performed  in  example  3.  There  is  another  method,  often  practised, 
erroneously  called  borrowing  ten,  —  explain  the  principle  on  which  it  is 
done.  When  we  subtract  units  from  units,  of  what  name  will  the 
remainder  be?  tens  from  tens,  what?  hundreds  from  hundreds,  what? 


1F  18.  SUBTRACTION  OF  SIMPLE  NUMBERS.  27 

NOTE.  —  It  has  been  usual  to  perform  subtraction,  where  the  figure 
in  the  subtrahend  is  larger  than  the  figure  above  it,  on  another  prin 
ciple.  If  to  two  unequal  numbers  the  same  number  be  added,  the  dif 
ference  between  them  will  remain  the  same.  Thus,  the  difference 
between  17  and  8  is  9,  and  the  difference  between  27  and  18,  each 
being  increased  by  10,  is  also  9.  Take  the  last  example. 

TENS.  UNITS.         Adding  10  units  to  5  units  in  the  minuend,  and  1 

£  ten  to  2  tens  in  the  subtrahend,  we  have  increased 

_  •  both  by  the  same  number,  and  the  remainder  is  not 

f.          g  altered,  being  58. 

This  method,  which  has  been  erroneously  called  borrowing  ten, 
may  be  practised  by  those  who  prefer,  though  the  former  is  more 
simple  and  equally  convenient. 

3.  From  10000  subtract  9. 

OPERATION.          SOLUTION.  —  In  this  example  we  have  0  units  from 

10000   '      w^ich  to  subtract  9  units,  and  going  to  tens  of  the 

minuend,  we  have  0  tens,  nor  hundreds,  nor  thousands  ; 

but  we  have  1  ten  thousand  from  which,  borrowing  10 

-         units,  we  have  9990,  that  is,  9  thousands,  9  hundreds 

9991          and  9  tens  left.     Taking  9  units  from  10  units,  we 

have  1  unit,  then  no  tens  in  the  subtrahend  from  9  tens 

in  the  minuend  leave  9  tens,  no  hundreds  from  9  hundreds  leave  9 

hundreds,  no  thousands  from  9  thousands  leave  9  thousands. 

4.  A  man  borrowed  713  dollars  and  paid  475  dollars  ;  how 
much  did  he  then  owe  ?  Ans.  238  dollars. 

5.  From  1402003  take  681404.  Rem.  720599. 

6.  What  is  the  difference  between  36070324301  and  280- 
40373315  ?  Ans.  8029950986. 

7.  From  81324036521  take  2546057867. 

Rem.  78777978654. 


To  PROVE  ADDITION  AND  SUBTRACTION.  —  Addition 
and  subtraction  are  the  reverse  of  each  other.  Addition  is 
putting  together  ;  subtraction  is  taking  asunder. 

1.  A  man  bought  40  sheep  2.  A  man  sold  18  sheep 

and  sold  18  of  them  ;  how  and  had  22  left  ;  how  many 

many  had  he  left  ?  had  he  at  first  ? 

40  —  18  =  22  sheep  left.  18  -|-  22  =  40  sheep  at  first. 

Ans.  Ans. 

Hence,  subtraction  may  be  proved  by  addition,  and  addition 
by  subtraction. 

To  prove  subtraction,  add  the  remainder  to  the  subtrahend, 
and,  if  the  work  is  right,  the  amount  will  be  equal  to  the 
minuend. 


28  SUBTRACTION  OF  SIMPLE  NUMBERS.  ^T  18 

To  prove  addition,  subtract,  successively,  from  the  amount, 
the  several  numbers  which  were  added  to  produce  it,  and,  if 
the  work  is  right,  there  will  be  no  remainder.  Thus  7  +  8 
_i_  6  =  21  ;  proof,  21  —  6  =  15,  and  15  —  8  =  7,  and  7  — 
7  =  0. 


.  —  Proof  by  excess  of  nines.  We  may  cast  out  the  nines  in 
the  remainder  and  subtrahend  ;  if  the  excess  equals  the  excess  found 
by  casting  out  the  nines  from  the  minuend,  the  work  is  presumed  to 
be  right. 

From  the  remarks  and  illustrations  now  given,  we  deduce 
the  following 

RULE. 

I.  Write  down  the  numbers,  the  less  under  the  greater, 
placing  units  under  units,  tens  under  tens,  &c.,  and  draw  a 
line  under  them. 

II.  Beginning  with  units,  take  successively  each  figure  in 
the  lower  number  from  the  figure  over  it,  and  write  the  re 
mainder  directly  below. 

III.  When  a  figure  of  the  subtrahend  exceeds  the  figure 
of  the  minuend  over  it,  borrow  1  from  the  next  left  hand 
figure  of  the  minuend  ;  and  add  it  to  this  upper  figure  as  10, 
in  which  case  the  left  hand  figure  of  the  minuend  must  be 
considered  one  less. 

NOTE.  —  Or  when  the  lower  figure  is  greater  than  the  one  above 
it  we  may  add  10  to  the  upper  figure,  and  1  to  the  next  lower  figure. 

EXAMPLES  FOR  PRACTICE. 

1.  If  a  farm  and  the  buildings  on  it  be  valued  at  10000, 
and  the  buildings  alone  be  valued  at  4567  dollars,  what  is  the 
value  of  the  land  ?  Ans.  5433  dollars. 

2.  The  population  of  New  York  in  1830  was  1,918,608; 
in  1840  it  was  2,428,921;    what  was  the  increase  in  ten 
years  ?  Ans.  510,313. 

3.  George  Washington  was  born  in  the  year  1732,  and 
died  in  the  year  1799  ;  to  what  age  did  he  live  ? 

Ans.  67  years. 

4.  The  Declaration  of  Independence  was  published  July 
4th,  1776;  how  many  years  to  July  4th  the  present  year? 

Questions.  —  If  18.  Addition  is  the  reverse  of  what?  Subtrac 
tion,  of  what  ?  How  will  you  show  that  they  are  so  ?  How  do  you 
prove  subtraction  ?  How  can  you  prove  addition  by  subtraction  ?  R» 
peat  the  rule  for  subtraction. 


H  19.  SUBTRACTION  OF  SIMPLE  NUMBERS.  29 

5.  The  Rocky  Mountains,  in  N.  A.,  are  12,500  feet  above 
the  level  of  the  ocean,  and  the  Andes,  in  S.  A.,  are  21,440 
feet ;  how  many  feet  higher  are  the  Andes  than  the  Rocky 
Mountains  ?  Ans.  8,940  feet. 

NOTE.  —  Let  the  pupil  be  required  to  prove  the  following  examples. 

6.  What  is  the  difference  between  7,648,203  and  928,671  ? 

Am.  6,719,532. 

7.  How  much  must  you  add  to  358,642  to  make  1,487,945  ? 

Ans.  1,129,303. 

8.  A  man  bought  an  estate  for  13,682  dollars,  and  sold  it 
again  for  15,293  dollars ;  did  he  gain  or  lose  by  it  ?  and  how 
much?  Ans.  1,611  dollars. 

9.  From  364,710,825,193  take  27,940,386,574. 

10.  From  831,025.403,270  take  651,308,604,782. 

11.  From  127,388,047,216,843  take  978,654,827,352. 


IT  19.  Review  of  Subtraction. 

Questions. — What  is  subtraction?  What  is  the  rule?  What 
is  understood  by  borrowing  ten  ?  Of  what  is  subtraction  the  reverse  ? 
How  is  subtraction  proved  ?  How  is  addition  proved  by  subtraction  * 

.   • 
EXERCISES. 

1.  How  long  from  the  discovery  of  America  by  Columbus, 
in  1492,  to  this  present  year? 

2.  Supposing  a  man  to  have  been  born  in  the  year  1773, 
how  old  was  he  in  1847  ?  Ans.  74. 

3.  Supposing  a.man  to  have  been  80  years  old  in  the  year 
1846,  in  what  year  was  he  born  ?  Ans.  1766. 

4.  There  are  two  numbers,  whose  difference  is  8764;  the 
greater  number  is  1JC37  ;  I  demand  the  less.     Ans.  6923. 

5.  What  number  is  that  which,  taken  from  3794,  leaves 
865  ?  Ans.  2929. 

6.  What  number  is  that  to  which  if  you  add  789,  it  will 
become  6350  ?  Ans.  5561. 

7.  A  man  possessing  an  estate  of  twelve  thousand  dollars, 
gave  two  thousand  five  hundred  dollars  to  each  of  his  two 
daughters,  and  the  remainder  to  his  son;  what  was  his  son's 
share  ?  A?is.  7000  dollars. 

8.  From  seventeen  million   take  fifty-six  thousand,  and 
what  will  remain  ?  Ans.  16,944,000. 


30  SUBTRACTION  OF  SIMPLE  NUMBERS.  If  19 

9.  What  number,  together  with  these  three,  viz.,  1301, 
2561,  and  3120,  will  make  ten  thousand?  Ans.  3018. 

10.  A  man  bought  a  horse  for  one  hundred  and  fourteen 
dollars,  and  a  chaise  for  one  hundred  and  eighty-seven  dol 
lar*  ;  how  much  more  did  he  give  for  the  chaise  than  for  the 
Horse  ? 

11.  A  man  borrows  7  ten  dollar  bills  and  3  one  dollar  bills, 
and  pays  at  one  time  4  ten  dollar  bills  and  5  one  dollar  bills ; 
how  many  ten  dollar  bills  and  one  dollar  bills  must  he  aftei- 
wards  pay  to  cancel  the  debt  ? 

Ans.  2  ten  doll,  bills  and  8  one  doll. 

12.  The  greater  of  two  numbers  is  24,  and  the  less  is  16 ; 
what  is  their  difference  ? 

13.  The  greater  of  two  numbers  is  24,  and  their  difference 
8 ;  what  is  the  less  number  ? 

14.  The  sum  of  two  numbers  is  40,  the  less  is  16 ;  what 
is  the  greater? 

EXERCISES   IX  ADDITION  AXD  SUBTRACTION. 

15.  A  man  carried  his  produce  to  market ;    he  sold  his 
pork  for  45  dollars,  his  cheese  for  38  dollars,  and  his  butter 
for  29  dollars ;   he  received,  in  pay,  salt  to  the  value  of  17 
dollars,  10  dollars'  worth  of  sugar,  5  dollars'  worth  of  molasses, 
and  the  rest  in  money  ;  how  much  money  did  he  receive  ? 

Ans.  80  dollars. 

16.  A  boy  bought  a  sled  for  28  cents,  and  gave  14  cents  to 
have  it  repaired ;  he  sold  it  for  40  cents ;  did  he  gain  or  lose 
by  the  bargain  ?  and  how  much  ?         Ans.  He  lost  2  cents. 

17.  One  man  travels  67  miles  in  a  day,  another  man  fol 
lows  at  the  rate  of  42  miles  a  day ;  if  they  both  start  from 
the  same  place  at  the  same  time,  how  far  will  they  be  apart 

at  the  close  of  the  first  day  ?  of  the  second  ?  of  the 

third  ?  of  the  fourth  ?        Am.  To  the  last,  100  miles. 

18.  One  man  starts  from  Boston  Monday  morning,  and 
travels  at  the  rate  of  40  miles  a  day ;   another  starts  from  the 
same  place  Tuesday  morning,  and  follows  on  at  the  rate  of 
70  miles  a  day ;  how  far  are  they  apart  Tuesday  night  ? 

Ans.  10  miles. 

19.  A  man,  owing  379  dollars,  paid  at  one  time  47  dollars 
nt  another  time  84  dollars,  at  another  time  23  dollars,  and  at 
another  time  143  dollars ;  how  much  did  he  then  owe  ? 

Ans.  82  dollars. 

20.  Four  men  bought  a  lot  of  land  for  482  dollars  ;  the  first 
man  paid  274  dollars,  the  second  man  194  dollars  less  than 


If  20.          MULTIPLICATION  OP  SIMPLE  NUMBERS.  31 

the  first,  and  the  third  man  20  dollars  less  than  the  second ; 
how  much  did  the  second,  the  third,  and  the  fourth  man  pay  ? 

(  The  second  paid  80. 
Ans.  \  The  third  pajd  60. 
(  The  fourth  paid  68. 

21  Four  men  bought  a  horse  ;  the  first  man  paid  21  dol 
lars,  the  second  18  dollars,  the  third  13  dollars,  and  the  fourth 
as  much  as  the  other  three,  wanting  16  dollars ;  how  much 
did  the  fourth  man  pay  ?  and  what  did  the  horse  cost  ? 

Ans.  Fourth  man  paid  —  dolls. ;  horse  cost  88  dolls. 
22.    From  1,000,000  take  1,  and  what  remains  ?     ( See  IT  17 
Ex.  3.) 


MULTIPLICATION  OF  SIMPLE  NUMBERS. 


1.  If  one  orange  costs  5  cents,  how  many  cents 
must  I  give  for  2  oranges  ?  -  how  many  cents  for  3 
oranges  ?  -  for  4  oranges  ? 

2.  One  bushel  of  apples  cost  20  cents  ;  how  many  cents 
must  I  give  for  2  bushels  ?  -  for  3  bushels  ? 

3.  One  gallon  contains  4  quarts  ;   how  many  quarts  in  2 
gallons  ?  -  in  3  gallons  ?  -  in  4  gallons  ? 

4.  Three  men  bought  a  horse  ;  each  man  paid  23  dollars 
for  his  share  ;  how  many  dollars  did  the  horse  cost  them  ? 

5.  In  one  dollar  there  are  one  hundred  cents  ;  how  many 
cents  in  5  dollars  ? 

6.  How  much  will  4  pairs  of  shoes  cost  at  2  dollars  a  pair  ? 

7.  How  much  will  two  pounds  of  tea  cost  at  43  cents  a 
pound  ? 

8.  There  are  24  hours  in  one  day  ;   how  many  hours  in  2 
days  ?  -  in  3  days  ?  -  in  4  days  ?  -  in  7  days  ? 

9.  Six  boys  met  a  beggar,  and  gave  him  15  cents  each  ; 
how  many  cents  did  the  beggar  receive  ? 

In  this  example  we  have  15  cents  (the  number  which  each 
boy  gave  the  beggar)  to  be  repeated  6  times,  (as  many  times 
as  there  were  boys.) 

When  questions  occur  where  the  same  number  is  to  be 
repeated  several  times,  the  operation  may  be  shortened  by  a 
rule  called  Multiplication. 

In  multiplication  the  number  to  be  repeated  is  called  the 
Multiplicand. 

The  number  which  shows  how  many  times  the  multiplicand 
vs  to  be  repeated,  is  called  the  Multiplier. 


32  MULTIPLICATION  OF  SIMPLE  NUMBERS.  11  20 

The  result,  or  answer,  is  called  the  Product. 

The  multiplicand  and  multiplier  taken  together  are  called 
Factors,  or  producers,  because  when  multiplied  together  they 
produce  -the  product. 

10.  There  is  an  orchard  in  which  are  5  rows  of  trees,  and 
27  trees  in  each  row  ;  how  many  trees  in  the  orchard  ? 

In  the  first  row,  .  .  .  27  trees.        SOLUTION.—  The  whole  nnm- 
"       second  27     "         her  of  trees  will  be  equal  to  the 

"       third  27     "         amount  of  Jive  27's  added  to- 


we  find  that  7 

taken  five  times  amounts  to  35. 
We  write  down  the  five  units, 

In  the  whole  orchard)  135  trees,    and  reserve  the  three  tens  ;  the 

amount  of  2  taken  five  times  is 

10,  and  the  3,  which  we  reserved,  makes  13,  which,  written  at  the 
left  of  units,  makes  the  whole  number  of  trees  135. 

If  we  have  learned  that  7  taken  5  times  amounts  to  35,  and  that  2 
taken  5  times  amounts  to  10,  it  is  plain  we  need  write  the  number 
27  but  once,  and  then,  setting  the  multiplier  under  it,  we  may  say, 

5  times  7  are  35,  writing 

Multiplicand,  27  trees  in  each  row.     down  the  5,  and  reserving 

Multiplier,  .     5  rows.  the  3  (tens)  as  in  addition. 

_  Again,  5  times  2  (tens)  are 

P^uct,   .    KStr*,,**  JLW,»(^ 

13,  (tens,)  as  before. 

From  the  above  example,  it  appears  that  multiplication  is 
a  short  way  of  performing  many  additions,  and  it  may  be 
defined,  —  The  method  of  repeating  one  of  two  numbers  as 
many  times  as  there  arc  units  in  the  other. 

SIGN:  —  Two  short  lines,  crossing  each  other  in  the  form 
of  the  letter  X,  are  the  sign  of  multiplication.  When  placed 
between  numbers  it  shows  that  they  are  to  be  multiplied 
together  ;  thus,  3  X  4  =  12,  signifies  that  3  times  4  are 
equal  to  12,  or  4  times  3  are  equal  to  12  ;  and  thus,  4  X  2  X 
7  =  56,  signifies  that  4  multiplied  by  2,  and  this  product  by 
7,  equals  56. 


Questions,  —  IF  20,  "When  questions  occur  in  which  the  same 
number  is  to  be  repeated  several  times,  how  may  the  operation  be  short 
ened  ?  In  multiplication,  what  is  the  multiplicand?  the  multiplier?  the 
product?  What  are  factors  ?  Why?  What  is  multiplication  ?  Illus 
trate  by  the  two  methods  of  performing  the  10th  example.  How  do 
you  define  multiplication  ?  What  is  the  sign  ?  Repeat  the  table. 


H20. 


MULTIPLICATION  OF  SIMPLE  NUMBERS. 


33 


NOTE.  —  Before  any  progress  can  be  made  in  this  rule,  the  follow 
ing  table  must  be  committed  perfectly  to  memory. 

MUI/TIPUCAT1OX   TABLE. 


2  times  0  are  0 

2X  1-=  2 

2x  2  =  4 
9  v  3  fi 

4x10  =  40 
4x  11  =  44 

4x  12  =  48 

7x    6=   42 
7  X    7  =   49 
7X    8=   56 
7  v    0          fi*3 

10  x    3=  30 
10  X    4=    40 
10  X    5=   50 
in  \'    fi        an 

**  /\  "  —  ^ 
2X  4=  8 
2x  5=10 
2x  6=12 
2x  7=14 
2x  8=16 
2X  9=18 
2  X  10  20 

5X  0=  0 
5X  1=  5 
5x  2=10 
5x  3  =  15 
5x  4  =  20 
5X  5  =  25 
.*»  v  fi  ^o 

7  X  10=    70 
7X  11=    77 
7x  12=   84 

10  x    7=  70 
10  X    8=   SO 
10  x    9=   90 
10  x  10=100 
10  x  11  =  110 
10  x  12  =  120 

Sx    0=     0 
SX    1=     8 
Sx    2=    16 
8x    3=   24 
Sx    4=   32 
SX    5—40 

2x  11  =  22 
2X12  =  24 

5x  7  =  35 

5x  8  =  40 

K  ^s  0  J/t 

11  X    0=     0 

iix  1=  11 

H\  s       O                 Of> 

3X  0=  0 
3X  1=  3 
3X  2=  6 
3x  3=  9 
3X  4=12 
3x  5=15 
3x  6=18 
3x  7  =  21 
3x  8  =  24 
3x  9  =  27 
3  X  10  =  30 
3XH=33 
3x  12  =  36 

5x10  =  50 
5X  11  =  55 
5X12  =  60 

Sx    6=   48 
8x    7=   56 
8x    8=   64 
8x    9=   72 
8x  10=  80 
8x  H=   8S 
8x  12=   96 

X    *  —   *"£ 
11  X    3=   33 
11  X    4=   44 
11  X    5=   55 
n^'    ft        fifi. 

6X  0=  0 
6x  1=  6 
6x  2=12 
6X  3=18 
6X  4  =  24 
6X  5  =  30 
6X  6  =  36 
6X  7  =  42 
6X  8  =  48 

fi  V  0  .^d 

11  x    7=    77 
11  X    8=   88 
11  X    9=   99 
11  x  10=310 
11  x  11  =  121 
11  x  12=132 

9x    0=     0 
9x    1=     9 
9x    2=    18 
9x    3=   27 
9x    4=   36 
9X5=   45 
9x    6=   54 
9x    7=   63 
9X    8=   72 
9x    9=   81 
9X10=   90 
9X  11=   99 
9x  12  =  108 

12  x    0=     0 
12  X    1=    12 

1  9  v     9      -    9d 

4x  0=  0 
4  X  1=4 
4X  2=  8 

A.  v  2  19 

6X  10  =  60 
6X  11  =  66 
6x  12  =  72 

It   X       *!  <^4 

12  X    3=   36 
12  X    4=   48 
12  X    5=   60 

19  \S     fi            *79 

4X  4=16 
4x  5  =  20 
4x  6  =  24 
4x  7  =  28 
4x  8  =  32 
4x  9  =  36 

7  X  0=  0 
7X  1=  7 
7X  2=14 
7x  3  =  21 
7X  4  =  28 
7X  5  =  35 

1^  X      O  i  £ 

12  X    7=   84 
12  X    8=   96 
12  X    9=108 
12  X  10=120 
12  X  11  =  132 
12  X  12=144 

10  X    0=     0 
10  X    1=    10 
10  X    2=   20 

34  MULTIPLICATION  OF  SIMPLE  NUMBERS. 

9x2  =  how  many  ?  4  X  3  X  2  =  24. 

4x6  =  how  many  ?  3x2x5  =  how  many  * 

8x9  =  how  many  ?  7xlX2  =  how  many  ? 

3x7  =  how  many  ?  8x3x2  =  how  many  ? 

5X5  =  how  many  ?  3x2x4x5  =  how  many  ? 


IT  SI.  1.  There  are  on  a  board,  3  rows  of  stars,  and  4 
stars  in  a  row  ;  how  many  stars  on  the  board  ? 

DIAGRAM  OF  STARS.         A  slight  inspection  of  the  diagram  will 

£     X     $:     :g         show  that  the  number  of  stars  may  be 

found  by  considering  that  there  are  either 

*     *     =fc     =fc         3  rows  of  4  stars  each,  (3  times  4  are  12,) 

X     $:     $:     :£         or  ^  rows  of  3  stars  each,  (4  times  3  are 

12;)  therefore,  we  may  use  either  of  the 

given  numbers  for  a  multiplier,  as  best  suits  our  convenience. 

We  generally  write  the  numbers  as  in  subtraction,  the  larger 

uppermost,  with  units  under  units,  tens  under  tens,  &c.    Thus, 

Multiplicand,  4  stars. 

Multiplier,      3  roivs.  NOTE.  —  4  and  3  are  the  factors, 

which  produce  the  product  12. 
Product,        12  stars,  Ans. 

This  diagram  of  stars  is  commended  to  the  particular  at 
tention  of  the  pupil,  as  it  is  intended  to  make  use  of  it  here 
after  in  illustrating  operations  in  multiplication  and  also  in 
division. 

First,  you  will  notice  the  terms  of  the  diagram,  and  their 
application. 

TERMS  OF  THE  DIAGRAM. 

(      Using  this  term  as  a  representation  or 
Stars  in  a  row.    <  symbol  of  the  multiplicand  and  one  factor 
(  of  the  product. 

Number  of  rows.  \  ,.  Y*™S  tV^S  **?  ^mbfol^f  the  ful~ 
(  tipiier  and  the  other  factor  of  the  product. 

I  Using  this  term  as  a  symbol  of  the  pro- 
duct,  for  when  the  stars  in  a  row  are  taken 
as  many  times  as  there  are  rows  of  stars, 
then  the  product  will  be  the  whole  number 
of  stars  contained  in  the  diagram. 

As  the  stars  in  a  row  symbolize  the  multiplicand,  it  follows 
that  the  multiplier  (number  of  rows)  in  reality  simply  express 
es  the  number  of  times  the  multiplicand  (stars  in  a  row)  is  to 


121.          MULTIPLICATION  OP  SIMPLE  NUMBERS.  36 

be  taken.  Hence,  to  multiply  by  1,  (1  row  of  stars,)  is  to  take 
the  multiplicand  (stars  in  a  row)  1  time  ;  to  multiply  by  2, 
(2  rows,)  is  to  take  the  multiplicand  (stars  in  a  row)  2  times ; 
to  multiply  by  3,  (3  rows,)  is  to  take  the  multiplicand  (stars 
in  a  row)  3  times,  and  so  on. 

ILLUSTRATION. — What  cost  7  yards  of  cloth  at  3  dollars  a 
yard?  (7  rows,  3  stars  in  a  row.)  The  two  numbers  as 
given  in  the  question  are  both  denominate  ;  but  how  are  they 
to  be  considered  in  the  operation  ?  The  price  of  7  yards  will 
evidently  be  7  times  the  price  of  1  yard,  that  is,  7  times  3 
dollars  ;  dollars  (number  of  stars)  is  the  thing  sought  by  the 
question  ;  and  hence,  3  dollars  being  of  the  same  name  as  the 
thing  or  answer  sought,  is  the  true  multiplicand.  That  num 
ber  which  was  yards  in  putting  the  question,  being  taken  for 
the  multiplier,  in  this  relation  is  not  to  be  considered  yards, 
but  times  of  taking  the  multiplicand  :  and  hence,  in  the  oper 
ation,  it  must  always  be  considered  an  abstract  number 
For  multiplication  is  a  short  way  of  performing  many  ad 
ditions,  and  to  talk  of  adding  3  dollars  to  itself  7  yards  times 
is  nonsense.  But  we  can  repeat  3  dollars  as  many  times  as 
1  yard  is  repeated  to  make  7  yards. 

There  is  then  a  true  multiplicand  and  a  true  multiplier. 
The  true  multiplicand  is  that  number  which  is  of  the  same 
name  as  the  answer  sought ;  the  true  multiplier  is  that  num 
ber  which  indicates  the  times  the  true  multiplicand  is  to  be 
repeated,  or  taken ;  but  as  it  respects  the  operation,  it  has 
been  shown  above  that  we  may  use  either  of  the  given  num 
bers  as  the  multiplier  ;  that  is,  the  multiplicand  and  multiplier 
may  change  places ;  still,  the  product  will  always  be  of  the 
same  name  as  the  true  multiplicand. 

This  application  of  the  terms  of  the  diagram  to  the  terms 
of  the  question  we  shall  call  symbolizing  the  question. 

Questions.  —  If  21.  You  have  in  your  book  a  diagram  of  stars; 
what  is  the  first  use  made  of  it  ?  What  is  the  difference  between  4  times 
7,  and  7  times  4  ?  Which  of  the  given  numbers  may  be  used  for  the 
multiplier  ?  What  are  the  terms  of  the  diagram  ?  What  do  these  terms 
symbolize  ?  6  times  7  are  42,  —  which  of  these  numbers  is  the  multipli 
cand?  the  multiplier?  the  product?  and  what,  in  the  diagram,  is  a  sym 
bol  of  each  ?  What  does  the  multiplier  express  ?  Show  by  the  diagram 
what  it  is  to  multiply  by  1,  by  2,  by  3,  &c.  What  must  the  multiplier 
always  be  considered?  What  do  you  understand  by  the  true  multipli 
cand?  by  the  true  multiplier?  What  will  the  product  always  be  i  Give 
an  example  to  show  that  you  understand  these  principle's.  What  do  you 
understand  by  symbolizing  a  question  ? 


36  MULTIPLICATION  OP  SIMPLE  NUMBERS.  IT  22. 

NOTE.  —  Let  the  teacher  see  to  it  that  these  principles  are  well 
understood  by  the  pupil  before  he  proceeds. 

As  the  pupil  advances,  the  teacher  should,  from  time  to  time,  refer 
him  back  to  a  review  of  these  principles. 


1.    What  will  84  barrels  of  flour  cost,  at  7  dollars 
a  barrel  ? 

SOLUTION.  —  The  price  of  84  barrels  will  evidently  be  84  times  the 
price  of  1  barrel,  7  stars  in  one  row  X  by  84,  number  of  rows,  7  dol 
lars  is  the  true  multiplicand,  &c.  ;  but  as  it  will  be  more  convenient, 
the  multiplicand  and  multiplier  may  change  places,  and  we  may 
consider  it  7  rows  of  84  stars  in  a  row,  and  multiply  the  number  of 
barrels,  84,  by  the  price  of  1  barrel,  thus  — 

Writing  the  larger  number  uppermost, 

OPEEATION.  as  *n  subtraction,  (^f  16,)  and  the  multi- 

Multiplicand,  84  P^er  un^er  ur'i;s  °f  the  multiplicand,  we 

Multiplier      '    7  be£in   at  the   ri»ht    hand  and  say,  7  X  4 

UipLl  'r'  _  [  (units)  =  28  (units)  =  2  tens  and  8  units  ; 

„  we  set  down  the  8.  units  in  units'  place,  as 

Product,         5b8  dolls.   in  addition,  and  reserving  the  2  tens,  we 

say,  7  X  8  (tens)  =  56  (tens,)  and  2  (tens) 

which  we  reserved,  make  58  (tens,)  or  five  hundred  and  8  tens,  which 
we  set  down  at  the  left  of  the  8  units,  and  the  whole  make  588  dol 
lars,  the  cost  of  84  barrels  of  flour,  at  7  dollars  a  barrel,  Ans. 

2.  A  merchant  bought  273  hats,  (stars  in  a  row,)  at  8  dol 
lars  each,  (number  of  rows  ;)  what  did  they  cost  (number  of 
stars)  ?  Ans.  2184  dollars.  ' 

3.  How  many  inches  are  there   in   253  feet,  (stars  in  a 
row,)  every  foot  being  12  inches  (number  of  rows)  ? 

OPERATION.    SOLUTION.—  The  product  of  12,  with  each  of  the 

10     s^on^cant  figures  or  digits,  having  been  committed  to 

memory  from  the  multiplication    -bL-,  it  is  just  as  easy 

-     to  multiply  by  12  as  by  a  single  figure.     Thus,   12 

Ans.  3036     times  3  are  30,  &c. 

4.  What  will  476  barrels  of  fish  cost,  at  11  dollars  a  bar 
rel  ?  Ans.  5236  dollars. 

From  these  examples  we  deduce  the  following 


_  Questlons«  — 1F22.  How  wil1  >'ou  exPlain  tne  first  exa.nple? 
When  you  multiply  units  by  units,  what  is  your  product?  When  tens 
by  units,  what  ?  How  can  you  multiply  by  12  ?  Hotf  do  you  write 
down  numbers  for  multiplication  ?  How  do  you  perform  multiplication 
when  the  inultiplu  does  not  exceed  12  ? 


U23.  MULTIPLICATION  OF  SIMPLE  NUMBERS.  37 

RULE. 

I.  To  set  down  numbers  for  multiplication.     Write  down 
the  multiplicand,  under  which  write  the  multiplier,  setting 
units  under  units,  tens  under  tens,  &c. 

II.  To  perform  multiplication  when  the  multiplier  does  not 
exceed  12.     Begin  at  the  right  hand,  and  multiply  each  figure 
in  the  multiplicand  by  the  multiplier,  setting  down  and  car 
rying  .as  in  addition. 

EXAMPLES    FOR    PRACTICE. 

•5.  A  farmer  sold  29  bags  of  wheat,  each  bag  containing  3 
bushels;  how  many  bushels  did  he  sell?  29  X  3  =  how 
many  ? 

6.  A  farmer,  who  had  two  farms,  raised  361  bushels  of 
wheat  on  one,  and  5  times  as  much  on  the  other ;  how  many 
bushels  did  he  raise  on  both  ?  Ans.  2166  bushels-. 

7.  A  miller  sold  42  loads  of  flour,  each  load  containing  9 
barrels,  at  7  dollars  a  barrel ;  how  many  barrels  of  flour  did 
he  sell,  and  what  did  the  whole  cost? 

Ans.  He  sold barrels  ;  cost,  2646  dollars. 

IT  S3.  1.  A  piece  of  valuable  land,  containing  33  acres, 
(number  of  rows,)  was  sold  for  246  dollars  an  acre,  (stars  in  a 
row  ;)  what  did  the  whole  cost  ? 

NOTE  1.  —  When  the  multiplier  exceeds  12,  it  is  more  convenient 
to  multiply  by  each  figure  separately  :  — 

FIRST  OPERATION.  SOLUTION. — In 

Multiplicand,  246  dollars,  price  of  1  acre,      this  example,   the 

Multiplier,         33  number  of  acres.  multiplier   consists 

of    3   tens    and    3 

Is,,  product,      739  dollars,  price  of  3  acres. 
units,  gives  us  738  dollars,  the  price  of  3  acres. 

Having  found  the  price  of  3  acres,  our  next  step  is  to  get 
the  price  of  30  acres. 

SECOND  OPEHATION.  To  do  this,  we  multiply  by  the 

246  dollars,  price  of  I  acre.        3  tens>   (thirty,)  and  write  the 

33  number  of  acres.  first  fi£ure  of  the  Product  (8)  *« 

tens'1  place,  that  is,  directly  under 

733  dollars,  price  0/3  acres.       the  figure  by  which  we  multiply. 
738    (tens)     price  o/30  acres.     For  the  price  of  30  acres  being 

10  times  the  price  of  3  acres,  it 

8118  dollars,  price  of  33  acres,     will  consist  of  the  same  figures, 

each  being  removed  1  place  to- 

waids  the  left,  by  which  its  value  is  increased  10  times.     Then  add- 
4 


38  MULTIPLICATION  OF  SIMPLE  NUMBERS.  IT  23 

ing  together  the  price  of  3  acres,  and  the  price  of  30  acres,  we  have 
the  price  of  33  acres. 

NOTE. — The  correctness  of  the  above  operation  results  from  the 
fact  that  when  units  (1st  order)  are  multiplied  by  units,  (1st  orde;,) 
the  product  is  units,  1st  order.  Tens  (2d  order)  X  units.  (1st  order,) 
the  product  is  units  of  the  2d  order.  Hundreds  (3d  order)  X  tens, 
(2d  order,)  the  product  is  units  of  the  4th  order. 

And  universally, — 

If  a  figure  of  any  order  be  multiplied  by  some  figure  of  an 
other  order,  the  product  will  be  units  of  that  order  indicated 
by  the  su?n  of  their  indices,  minus  1.  Thus,  7  of  the  5th 
order,  (70000,)  multiplied  by  4  of  the  3d  order,  (400,)  their 
indices  being  5  +  3  =  8,  and  8  —  1  =  7,  their  product  will 
be  28  units  of  the  7th  order,  that  is,  28  millions. 

2.  How  many  yards  in  23  pieces  of  broadcloth,  each  piece 
containing  67  yards? 

OPERATION.  SOLUTION.  —  Multiplying  67   yards 

67  yards  in  each  piece,     by  3,  we  get  201  yards  in  3  pieces  ;  and 

23  number  of  pieces,         multiplying  67  by  2  tens,  we  get  134 

tens,  =  1340  yards  in  20  pieces.     Add 

201  yards  in  3  pieces.  the  two  products  together,  and  we  get 
134  "  "  20  pieces.  1540  yards  (No.  of  stars)  in  23  pieces. 

Ans.   1540  yards. 
1541      "      "  23  pieces. 

Hence, —  To  perform  multiplication  when  the  multiplier 
exceeds  12, — 

RULE. 

I.  Multiply  the  multiplicand  by  each  figure  in  the  multi 
plier  separately,  first  by  the  units,  then  by  the  tens,  &c.,  re 
membering  always  to  place  the  first  figure  of  each  product 
directly  under  its  multiplier. 

II.  Having  multiplied  in  this  manner  by  each  figure  in  the 
multiplier,  add  these  several  products  together,  and  their  sum 
will  be  the  answer. 

PROOF.  —  Take  the  multiplicand  for  the  multiplier,  and  the  multi 
plier  for  the  multiplicand,  and  if  the  product  be  the  same  as  at  first, 
the  work  may  be  supposed  to  be  right. 

Questions.  —  f  23.  How  do  you  multiply  when  the  multiplier 
exceeds  12?  When  do  you  write  the  first  figure  of  each  product? 
Why  ?  What  do  you  do  with  the  several  products?  Repeat  the  rule. 
VV  hat  is  the  method  of  proof?  A  figure  of  any  one  order  multiplied  by 
some  figure  of  another  order,  the  product  will  be  what » 


MULTIPLICATION  OP  SIMPLE  NUMBERS.  39 

EXAMPLES   FOR    PRACTICE. 

3.  There  are   320  rods  in  a  mile  ;  how  many  rods   are 
there  in  57  miles  ?     320  X  57  =  how  many  ? 

4.  It  is  436  miles  from  Boston  to  the  city  of  Washington  ; 
how  many  rods  is  it  ? 

5.  What  will  784  chests  of  tea  cost,  at  69  dollars  a  chest  ? 
784  X  69  =  how  many  ? 

6.  If  1851  men  receive  758  dollars  apiece,  how  many  dol 
lars  will  they  all  receive  ?  Am.  1403058  dollars. 

NOTE.  — Proof  by  the  excess  of  nines.    Casting  out  the  nines  in  the 

multiplicand,  we  have  an  excess  of  6,  which  we  write  he- 

1851     fore  the  sign  of  multiplication.     Also,  we  find  the  excess 

758     in  the  multiplier  to  be  2,  which  we  write  after  the  sign. 

The  product  of  the  nines  cast  out  from  each  factor  will  be 

1403058     an  exact  number  of  nines,  since  every  nine  multiplied  by 

nine  produces  an  exact  number  of  nines.     Hence,  if  there 

PROOF.      is  an  excess  of  nines  in  the  entire  product,  it  must  be  from 

3  an  excess  in  the  product  of  the  excesses,  6  and  2,  found 

6X2       in  the  factors.     Multiplying  6  by  2,  and  casting  out  nine 

3  from  the  product,  we  write  the  excess,  3,  over  the  sign; 

and  casting  out  the  nines  from  the  product  of  the  factors, 

we  find  the  excess  will  be  the  same  number  3,  which  we  write  under 

the  sign,  and  presume  that  the  work  is  right. 

7.  There  are  24  hours  in  a  day  ;  if  a  ship  sail  7  miles  in 
an  hour,  how  many  miles  will  she  sail  in  1  day,  at  that  rate  ? 
how  many  miles  in  36  days  ?  how  many  miles  in  1  year,  or 
365  days  ?  Ans.  61320  miles  in  1  year. 

8.  A  merchant  bought  13  pieces  of  cloth,  each  piece  con 
taining  28  yards,  at  6  dollars  a  yard ;  how  many  yards  were 
there,  and  what  was  the  whole  cost? 

Ans.  to  the  last,  2184  dollars. 
9.   Multiply  37864  by    235.         Product,.      8898040. 

10.  «         29831    "     952.  "  28399112. 

11.  "         93956    «  8704.  "         817793024. 

12.  The  factors  of  a  certain  number  are  25  and  87  ;  what 
is  the  number?  Ans.  2175. 

13.  A  hatter  sold  15  cases  of  hats,  each  containing  24  hats 
worth  8  dollars   apiece ;  how  many  hats  did  he   sell,  and  to 
how  many  dollars  did  they  amount  ? 

A?is.  to  the  last,  2880  dollars. 

14  A  grazier  sold  23  head  of  cattle  every  year  for  6  years, 
at  an  average  price  of  17  dollars  a  head ;  how  many  head  of 
cattle  did  he  sell,  to  how  much  did  they  amount  each  year, 
and  to  how  much  did  they  amount  in  6  years  ? 

Ans.  to  the  last,  2346  dollars. 


40  MULTIPLICATION   OF  SIMPLE  NUMBERS.  fl  24 

Contractions  in  Multiplication. 

5T  24.     I.    When  the  multiplier  is  a  composite  number. 

Any  number  which  can  be  produced  by  multiplying-  two  or 
more  numbers  together,  is  called  a  Composite  number,  and 

The  numbers  which  are  multiplied  together  to  produce  it 
are  called  its  Component  parts,  or  Factors  ;  thus,  15  can  be 
produced  by  multiplying  together  3  and  5,  and  is,  therefore,  a 
composite  number,  and  the  numbers  3  and  5  are  its  compo 
nent  parts. 

So,  also,  24  is  a  composite  number.  Its  component  parts 
may  be  2  and  12,  (2  X  12  =  24,)  or  3  and  8.  (3  X  8  =  24,) 
or  4  and  6,  (4  x  6  =  24,)  or  2,  3,  and  4,  (2  X  3  X  4  =  24,) 
or  2,  2,  2,  3,  (2  X  2  X  2  X  3  =  24.) 

1.  What  will  18  yards  of  cloth  cost,  at  4  dollars  a  yard  ? 
3x6=  18.  It  follows,  therefore,  that  3  and  6  are  compo 
nent  parts  of  18. 

OPERATION.  SOLUTION.  —  If  1  yard  cost  4  dol- 

4  dollars,  cost  of  1  yard.        l&rs,  3  yards  will  cost  3  times  4  dol- 

3  yards.  lars>  =  12  dollars  ;  and,  if  3  yards 

cost  12  dollars,  18  yards  (3X6  = 

12  dollars,  cost  of  3  yards.       18)  will  cost  G  times  as  much  as  3 
6  (3  X  6  =)  13  yards.          yards,  that,  is,  6  times  12  dollars  = 

72  dollars.     Hence, 
72  dollars,  cost  of  13  yards. 

To  perform  multiplication  when  the  multiplier  exceeds  12, 
and  is  a  composite  number,  — 


I.    Separate  the  multiplier  into  two  or  more  component 
parts,  or  factors. 

Multiply  the  multiplicand  by  one  of  the  component 
parts,  and  the  product  thus  obtained  by  the  other,  and  so  on, 
if  the  component  parts  be  more  than  two,  till  vou  have  multi 
plied  by  each  one  of  them.  ,  The  last  product  will  be  the 
product  required. 

Questions.  —  «j[  24.     What  is  a  composite  number?    What  are  the 
component  parts,  or  factors?     Why  is  15  a  composite  number  ?     How 
any  factors  maj  a  composite  number  have  ?     Is  11  a  composite  nurn- 
v\  ay  not  s     Explain  the  1st  example.     How  do  you  multiply  by 
a  composite  number?      Does  it  matter  by  which  factor  you   multiply 
y°U  perforniea  the  3  ^rations,  (Ex.  2.)  and  compared  the*/ 


MULTIPLICATION   OF  SIMPLE  NUMBERS,  4 

EXAMPLES    FOR    PRACTICE. 

2.  What  will  136  tons  of  potash  cost,  at  96  dollars  per 
ton? 

Let  the  pupil  make  3  operations,  and  multiply,  1st  by  12 
and  8 ;  2dly,  by  4,  4,  and  6  ;  3dly,  by  96,  and  compare  the 
operations ;  he  will  find  the  results  to  be  the  same  in  each 
case.  Ans.  13056  dollars. 

3.  Supposing  342  men  to  be  employed  in  a  certain  piece 
of  work,  for  which  they  are  to  receive  112  dollars  each;  how 
much  will  they  all  receive  ? 

8  X  7  X  2=  112.  Ans.  38304  dollars. 

4.  How  many  acres  of  land  in  48  farms,  each  containing 
367  acres?  Ans.   17616  acres. 

5.  Supposing  168  persons  to  be  employed  in  a  woollen 
factory,  at  an  average  price  of  274  dollars  each  per  year ;  how 
much  will  they  all  receive  ? 

8x7x3  =  168.  Ans.  46,032  dollars. 

6.  Multiply          853  by  56,          Product,          47,768. 

7.  »  18109   «    35.  «  633,815. 

8.  »         1947271    «    81.  "        157,728,951. 

1T  25.     II.    When  the  multiplier  is  10,  100,  1000,  #c. 

1.    What  will  10  acres  of  land  cost,  at  25  dollars  per  acre  ? 

SOLUTION.  —  The  price  of  10  acres  will  be  10  times  the  price  of  1 
acre,  or  10  times  25  dollars. 

Now  if  we  annex  a  cipher  to 

25,  the  5  units  are,  made  5  tens, 

25  dollars,  price  of  I  acre.          Aftd  the  2  tells  are  made  2  hun. 

250  dollars,  price  o/lO  acres,     dreds.     Each  figure,  then,  is  in 
creased  ten-fold,  and  consequently 
the  whole  number  is  multiplied  by  10. 

It  is  also  evident  that  if  2  ciphers  were  annexed  to  25,  the  5  units 
would  be  made  5  hundreds,  and  2  teas  would  be  made  2  thousands, 
each  figure  being  increased  a  hundred  fold,  or  multiplied  by  100.  If 
3  ciphers  were  annexed,  each  figure  would  be  multiplied  by  1000,  &c 
Hence, 

When  the  multiplier  is  10,  100,  1000,  or  1,  with  any  num 
ber  of  (.  iphei  s  a?inexed, — 

RULE. 

Annex  as  many  ciphers  to  the  multiplicand  as  there  are 

Questions.  —  If  25.  How  are  the  figures  of  a  number  affected,  by 
placing  one  cipher  at  the  right  hand?  two  ciphers?  three  ciphers?  &c. 
How,  then,  do  you  multiply  by  1,  with^ny  number  of  ciphers  annexed? 


42  MULTIPLICATION  OP  SIMPLE  NUMBERS.          IF  26 

ciphers  in  the  multiplier,  and  the  multiplicand,  so  increased, 
will  be  the  product  required. 

EXAMPLES    FOR    PRACTICE. 

2.  What  will  76  barrels  of  flour  cost,  at  10  dollars  a  bar 
rel  ?  Am.  760  dollars. 

3.  If  100  men  receive  126  dollars  each,  how  many  dollars 
will  they  al .  receive  ?  Ans.   12600  dollars. 

4.  What  will   1000  pieces  of  broadcloth  cost,  estimating 
each  piece  at  312  dollars  ?  Ans.  312000  dollars. 

5.  Multiply    5682  by    10000. 

6.  «        82134   "    100000 

IT  26.     III.    When  there  are  ciphers  on  the  right  hand  of 
the  multiplicand,  multiplier,  either  or  both. 

1.    What  will  40  acres  of  land  cost,  at  27  dollars  per  acre  ? 

OPERATION.  SOLUTION.  — The  price  of  40 

27    dollars,  price  of  1  acre.        acres  will  be  40  times  the  price 
4  of  1  acre.     But  40  being  a  com 

posite   number,    (4X10  =  40,) 


ino     j  77  •        ^  A  we  multiply  by  4,  one   com  no- 

1  n«n  &"'  ^^  °ffi«CreS'       nent  part,  to  get  ihe  price  of  4 

JO  dollars,  price  0/40  acres,     acres,  and  then  to  multiply  the 

price  of  4  acres  by  10,  the  other 
component  part,  we  annex  a  cipher  to  get  the  price  of  40  acres. 

2.    What  will  200  acres  of  land  cost,  at  400  dollars  an 
acre  ? 

FIRST  OPERATION.  SOLUTION. — The  200  acres 

400  dollars,  price  of  1  acre.          will  cost  200  times  the  price 
200  of  1  acre.     We  see  in  the' op 

eration  that   the  product  is  8 
with  4   ciphers   at   the   right 

000  hand,  the  same  number  as  in 

800  the    multiplicand   and    multi 

plier  counted  together.      We 
80000  dollars,  price  of  200  acres,     may  then  shorten  the  opera- 

tion ,  as  follows  :  — 

SECOND  OPERATION.         Multiplying  the  significant  figures  together, 
400  we  place  their  product,  8,  under  the  2.     Then 

200  we  annex  to  this  product  4  ciphers,  the  num- 

ber  in  both  factors.     Hence, 
80000 

To  perform  multiplication  when  there  are  ciphers  on  the 
right  hand  of  either,  or  both^the  factors, — 


1T27.          MULTIPLICATION  OF  SIMPLE  NUMBERS.  4S 

RULE. 

I.  Set  the  significant  figures  under  each  other,  placing  the 
ciphers  at  the  right  hand. 

II.  Multiply  the  significant  figures  together. 

III.  Annex  as  many  ciphers  to  the  product  as  there  are  on 
the  right  hand  of  both  the  factors. 

EXAMPLES    FOR    PRACTICE. 

3.  If  1300  men  receive  460  dollars  apiece,  how  many  dol 
lars  will  they  all  receive  ?  Ans.  598000  dollars. 

4.  It  takes  200  shingles  to  lay  1  course  on  the  roof  of  a 
barn,  and  there  are  60  courses  on  each  of  the  two  sides ;  how 
many  shingles  will  it  take  to  cover  the  barn  ? 

Ans.  24000. 

5.  A  certain  storehouse  contains  30  bins  for  storing  wheat, 
and  each  bin  will  hold  400  bushels ;  how  many  bushels  of 
wheat  can  be  stored  in  it  ?  Ans.  12000  bushels. 

IT  27.  IV.  When  there  are  ciphers  between  the  significant 
figures  of  the  multiplier. 

1.    What  is  the  product  of  378,  multiplied  by  204? 

FIRST  OPERATION.          Multiplying  by  a  cipher    SECOND  OPERATION 

378               produces  nothing.  There-  378 

204               f°re>  in  tne  multiplication,  204 
we  may  omit  the  cipher, 

and  multiply  by  the  sig-  1512 

run"  cant  figures  only,  as  in  756 
the  second  operation. 

Hence,  to  perform  mul-  77112 
77112              tiplication, 

WJien  there  are  ciphers  between  the  significant  figures  of 
the  multiplier,  — 

RULE. 

Omit  the  ciphers,  and  multiply  by  the  significant  figures 
only,  remembering  to  place  the  1st  figure  of  each  product 
directly  under  its  multiplier. 

Questions.  —  Tf  26.  How  do  you  set  down  numbers  for  multipL- 
cation,  when  there  are  ciphers  on  the  right  hand  of  the  multiplicand, 
multiplier,  either  or  both?  How  do  you  multiply?  How  many  ciphers 
do  you  annex  to  the  product?  If  there  were  2  ciphers  on  the  right  hand 
of  your  multiplicand,  and  5  on  the  right  hand  of  your  multiplier,  how 
many  would  you  annex  to  the  product? 

IF  27»  When  there  are  ciphers  between  the  significant  figures  of  the 
mi^tiplier,  now  do  you  multiply?  Where  do  you  set  the  1st  figure  of 
the  product  ? 


44  MULTIPLICATION  OF  SIMPLE  NUMBERS.  IT  2& 

EXAMPLES    FOR    PRACTICE. 

2.  Multiply  154326  by  3007.         Product,  46405S2S2. 

3.  Multiply  543  by  206.  Product,  111858. 

4.  Multiply  1620  by  2103.  Product,  3406S60. 

5.  Multiply  36243  by  32004.        Product,  1159920972. 

6.  Multiply  101,010,101  by  1,001,001. 

Product,  101,111,212,111,101. 

V  98.    Other  Methods  of  Contraction. 

I.  When  the  multiplier  is  9,  99,  or  any  number  of&s,  — 

Annex  as  many  ciphers  to  the  multiplicand  as  there  are 
nines  in  the  multiplier,  and  from  the  number  thus  produced, 
subtract  the  multiplicand  ;  the  remainder  will  be  the  product. 
Thus, 

Multiply  6547  by  999. 

OPERATION. 

6547000  Let  the  pupil  prove  the  operation  by  actual  mul- 

6547  tiplication. 

6540453  Ans. 

II.  When  the  multiplier  is  13,  14,  15,  16,  17,  18,  or  19,— 
Multiply  32046375  by  14. 

Place  the  multiplier  at  the  right  of  the  mul 

tiplicand,  with  the  sign  of  multiplication   be- 

32046375     X  14     tween  them  ;  multiply  the  multiplicand  by  the 

128185500  unit  figure  of  the  multiplier,  and  set  the  product 

one  place  to  the  right  of  the  multiplicand.    This 

448049250   A?is.     product,  added  to  the  multiplicand,  makes  the 

true  product. 


.—  If  the  multiplier  be  101,  102,  $c.,  to  109,— 
Multiply  72530486  by  103. 

OPERATION. 

72530486       X  103        Multiply  as  above,  and  set  the  product  two 
217591458  places  to  the  right  of  the  multiplicand,  and 

•74706400re  An*.      ^  them  t0ge'her  &r  the  tme  pr°dUCt- 

Questions.  —  «[[  28.  When  the  multiplier  is  9,  99,  &c.,  why  does 
the  contraction,  as  above  directed,  give  the  true  product?  AHS.  Multi 
plying  by  9  repeats  the  multiplicand  9  times  ;  annexing  a  cipher  repeats 
or  increases  it  10  times,  which  is  1  time  too  many  :  hence  the  rule,  sub 
tract  it  1  time,  &c.  When  the  multiplier  is  13,  14,  <5cc.,  why  ?  When 
101,  102,  &c.,  why  ?  When  21,  31,  &c.,  why? 


1T29.          MULTIPLICATION  OF  SIMPLE  NUMBERS*  45 

III.    When  the  multiplier  is  21,  31,  and  so  on  to  91, — 
Multiply  83107694  by  31. 

OPERATION.  Multiply  by  the  tens'1  figure  only  of  the  mul- 

83107694  X  31     tiplicand,  and  set  the  unit  figure  of  the  product 

249323082  under  the  place  of  the  tens,  and  so  on ;  then 

add  them  together  for  the  true  product. 
2576338514  Ans. 


IF  29.    Review  of  Multiplication. 

Questions. — What  is  multiplication,  and  how  defined?  Explain 
the  use  of  the  diagram  of  stars,  and  show  its  application  to  Ex.  1, 
1f  22.  What  must  the  true  multiplier  always  be?  the  product?  Why 
can  the  factors  exchange  places?  How  do  you  multiply  by  12,  or  less? 
by  a  number  greater  than  12?  by  a  composite  number?  by  1  with 
ciphers  annexed?  by  any  number  with  ciphers  annexed?  when  there 
are  ciphers  between  the  significant  figures  ?  When  units  of  different 
orders  are  multiplied  together,  of  what  order  is  the  product  ? 

EXERCISES. 

1.  An  army  of  10700  men,  having  plundered  a  city,  took 
so  much  money,  that,  when  it  was  shared  among  them,  each 
man  received  46  dollars ;  what  was  the  sum  of  money  taken  ? 

Ans.  492200  dollars. 

2.  Supposing  the  number  of  houses  in  a  certain  town  to 
be  145,  each  house,  on  an  average,  containing  two  families, 
and  each  family  6  members,  what  would  be  the  number  of 
inhabitants  in  that  town  ?  Ans.   1740. 

3.  If  46  men  can  do  a  piece  of  work  in  60  days,  how  many 
men  will  it  take  to  do  it  in  one  day  ?  Ans.  2760. 

4.  Two  men  depart  from  the  same  place,  and  travel  in  op- 
posi'e  directions,  one  at  the  rate  of  27  miles  a  day,  the  other 
31  miles  a  day ;  how  far  apart  will  they  be  at  the   end  of  6 
days  ?  Arts.  348  miles. 

5.  What  number  is  that,  the  factors  of  which  are  4,  7,  6 
and  20  ?  A?is.  3360. 

6.  If  18  men  can  do  a  piece  of  work  in  90  days,  how  long 
will  it  take  one  man  to  do  the  same  ?          Ans.  1620  days. 

7.  What  sum  of  money  must  be  divided  between  27  men, 
so  that  each  man  may  receive  115  dollars  ? 

8.  There  is  a  certain  number,  the  factors  of  which  are  89 
and  265 ;  what  is  that  number  ? 

9.  What  is  that  number,  of  which  9,  12,  and  14  are  fac 
tors? 

10.  If  a  carriage  wheel  turn  round  346  times  in  running 


46  MULTIPLICATION  OP  SIMPLE  NUMBERS.  IT  29 

1  mile,  how  many  times  will  it  turn  round  in  the  distance 
from  New  York  to  Philadelphia,  it  being  95  miles  ? 

Ans.  32870. 

11.  In  one  minute  are  60  seconds,  how  many  seconds  in 

4  minutes  ?  in  5  minutes  ?  in  20  minutes  ?  in 

40  minutes  ?  Ans.  to  the  last,  2400  seconds. 

12.  In  one  hour  are  60  minutes ;  how  many  seconds  in  an 

nour  ?    in  two  hours  ?    how  many  seconds  from  nine 

o'clock  in  the  morning  till  noon  ? 

Ans.  to  the  last,  10800  seconds. 

13.  Multiply  275827  by  19725.    Product,  5440687575. 

14.  Two  men,  A  and  B,  start  from  the  same  place  at  the 
same  time,  and  travel  the  same  way ;  A  travels  52  miles  a 
day,  and  B  44  miles  a  day ;  how  far  apart  will  they  be  at  the 
end  of  10  days  ?  Ans.  80  miles. 

15.  A  farmer  sold  468  pounds  of  pork  at  6  cents  a  pound, 
and  48  pounds  of  cheese  at  7  cents  a  pound,  and  received  in 
payment  42  pounds  of  sugar  at  9  cents  a  pound,  100  pounds 
of  nails  at  6  cents  a  pound,  108  yards  of  sheeting  at  10  cents 
a  yard,  and  12  pounds  of  tea  at  95  cents  a  pound  ;  how  many 
cents  did  he  owe  ?  Ans.  54  cents. 

16.  A  boy  bought  10  oranges  ;  he  kept  7  of  them,  and  sold 
the  others  for  5  cents  apiece  ;  how  many  cents  did  he  receive  ? 

Ans.  15  cents. 

17.  The  component  parts  of  a  certain  number  are  4,  5,  7, 
6,  9,  8,  and  3  ;  what  is  the  number  ?  Ans.  181440. 

18.  In  1  hogshead  are  63  gallons ;  how  many  gallons  in  8 
hogsheads  ?     In  1  gallon  are  4  quarts ;  how  many  quarts  in 
8  hogsheads  ?     In  1  quart  are  2  pints  ;  how  many  pints  in  8 
hogsheads  ?  Ans.  to  the  last,  4032  pints. 

19.  The  component  parts  of  a  multiplier  are  5,  3  and  5, 
and  the  multiplicand  is  118;    what  is  the  multiplier?   what 
the  product  ?  Ana.  to  the  last — the  product  is  8850. 

20.  An  army  consists  of  5  divisions,  each  division  of  8  bri 
gades,  each  brigade  of  4  regiments,  each  regiment  of  9  com 
panies,  and  each   company  of  77  men,  rank  and  file ;    the 
number  of  officers,  &c.,  to  the  whole  army  is  42,  the  number 
belonging  peculiarly  to  each  division  is  19,  to  each  brigade 
25,  to  each  regiment  11,  and  to  each  company  14  ;  how  many 
men  in  the  aroy  ?  A?is.  133937. 


H  30.  DIVISION   OF  SIMPLE  NUMBERS.  4V 


DIVISION   OF   SIMPLE  NUMBERS. 

If  SO.  1.  James  has  12  apples  in  a  basket,  which  he 
distributes  equally  among  several  boys,  giving  them  4  apples 
each  ;  how  many  boys  receive  them  ? 

SOLUTION.  —  He  can  give  the  apples  to  as  many  boys  as  the  times 
he  can  take  4  apples  out  of  the  basket,  which  is  3  times.  Ans.  3  boys. 

2.  If  a  man  travel  4  miles  in  an  hour,  in  how  many  hours 
will  he  travel  24  miles  ? 

SOLUTION. — It  will  take  him  as  many  hours  as  4  is  contained 
times  in  24.  Ans.  6  hours 

3.  James  divided  28  apples  equally  among  3  of  his  com 
panions  ;  how  many  did  he  give  to  each  ? 

SOLUTION.  —  The  28  apples  are  to  be  divided  into  3  equal  parts,  and 
one  part  given  to  each  boy,  who  will  thus  receive  9  apples.  It  will 
require  27  apples  to  give  3  boys  9  apples  each,  since  9X3  =  27. 
There  will  be  one  apple  left,  which  must  be  cut  into  3  equal  parts, 
and  1  part  given  to  each  boy. 

NOTE.  —  If  a  unit,  or  whole  thing,  be  divided  into  2  equal  parts,  one  of  those 
parts  is  called  one  half;  if  into  3  equal  parts,  1  part  is  called  1  third;  two 
parts  are  called  2  thirds,  &c.  If  divided  into  4  equal  parts,  one  part  is  called  1 
fourth,  or  one  quarter  ;  2  parts  are  called  2  fourths,  or  2  quarters ;  3  parts,  3 
fourths,  or  3  quarters,  &c.  If  divided  into  5  equal  parts,  me  parts  are  called 
fifths.  If  into  6  equal  parts,  the  parts  are  called  sixths,  &c. 

4.  Seven  men  bought  n  barrel  of  flour,  each  man  paying 
an  equal  share  ;  for  what  part  of  the  barrel  did  1  man  pay  ? 

2  men  ?     3  men  ?     4  men  ?     5  men  ? 

6  men  ? 

5.  Twelve  men  built  a  steamboat,  each  man  doing  an  equal 
share  of  the  work ;  how  much  of  the  work  did  3  men  do  ? 
5  men  ?    7  men  ?    9  men  ?    11  men  ? 

6.  A  boy  had  two  apples,  and  gave  one  half  an  apple  to 
each  of  his  companions ;  how  many  were  his  companions  ? 

Ans.  4. 

7.  A  boy  divided  four  apples  among  his  companions,  by 
giving  them  one  third  of  an  apple  each ;  among  how  many 
did  he  divide  his  apples  ?  Ans.  12. 

Questions.  — 1[30.  What  do  you  understand  by  1  half  of  any 
thing  or  number  ?  1  third  ?  2  thirds  ?  1  seventh  ?  4  sevenths  ?  6  fif 
teenths?  8  tenths?  5  twentieths?  9  twelfths?  How  many  halves  make 
a  whole  one  ?  How  many  thirds  ?  fourths  ?  sevenths  ?  ninths  ?  twelfths  T 
fifteenths  ?  twentieths  ?  &c.  How  many  thirds  make  three  whole  ones  ? 


48  DIVISION  OF  SIMPLE  NUMBERS,  If  31. 

8.  How  many  quarters  in  5  oranges  ? 

SOLUTION.  —  In  1  orange  there  are  4  quarters,  and  in  5  oranges 
there  are  5  X  4  ==  20  quarters.  Ans. 

9.  How  many  oranges  would  it  take  to  give  12  boys  one 
quarter  of  an  orange  each  ?  Ans.  3  or. 

10.  How  much  is  one  half  of  12  apples  ?         Ans.  6  ap. 

11.  How  much  is  one  third  of  12  ? 

12.  How  much  is  one  fourth  of  12?  Ans.  3. 

13.  A  man  had  30  sheep,  and  sold  one  fifth  of  them  ;  how 
many  of  them  did  he  sell  ?  Ans.  6. 

14.  A  man  purchased  sheep  for  7  dollars  apiece,  and  paid 
for  them  all  63  dollars  ;  what  was  their  number  ?     Ans.  9. 


1.    How  many  oranges,  at  3  cents  each,  may  be 
bought  for  12  cents  ? 

SOLUTION.  —  As  many  times  as  3  cents  can  be  taken  from  12  cents, 
BO  many  oranges  may  be  bought  ;  the  object,  therefore,  is  to  find  how 
many  times  three  is  contained  in  12. 

12  cents. 
First  orange,       3  cents. 

9  We  see,  in  this  example,  that  12  con- 

Second  orange,    3  cents.     tains  3  f9ur  times,  for  we  subtract  3 

from  12  four  times,  after  which  there  is 

no  remainder  ;    consequently,  subtrac- 

—   .    .  tion  alone  is  sufficient  for  the  operation  ; 

1  hira  orange,     3  cents.     but  we  may  come  to  the  same  result  by 

—  a  much  shorter  process,  called  division 

Ans.  4  oranges. 
Fourth  orange,    3  cents. 

0 

We  see  from  the  above,  that  one  number  will  be  contained 
In  another  as  many  times  as  it  can  be  subtracted  from  it,  ane1 
hence,  that 

Division  is  a  short  way  of  performing  many  subtractions  of 
the  same  number.  The  minuend  is  called  the  dividend,  the 
number  which  is  subtracted  at  one  time  is  called  the  divisor, 
and  the  number  which  indicates  the  number  of  times  the  sub 
traction  is  performed  is  called  the  quotient. 

The  cost  of  one  orange,  (3  cents,)  multiplied  by  the  number 
of  oranges,  (4,)  is  equal  to  the  cost  of  all  the  oranges,  (12 
cents  ;)  12  is,  therefore,  a  product,  and  3  one  of  its  factors  ; 
and  to  find  how  many  times  3  is  contained  in  12,  is  to  find 


DIVISION  OP  SIMPLE  NUMBERS.  49 

the  other  factor,  which,  multiplied  into  3,  will  produce  12. 
Hence,  the  process  of  division  consists  in  finding  one  factor 
of  a  product  when  the  other  is  known. 

2.  A  man  would  divide  12  cents  equally  among  3  children ; 
how  many  would  each  child  receive  ? 

SOLUTION.  —  The  numbers  in  this  are  the  same  as  in  the  former 
example,  but  the  object  is  different.  In  the  former  example,  the 
object  was  to  see  how  many  times  3  cents  are  contained  in  12  cents ; 
in  this,  to  divide  12  cents  into  3  equal  parts.  Still  the  object  is  to 
find  a  number,  which,  multiplied  into  3,  will  produce  12.  This,  as  in 
the  former  example,  is,  Ans.  4  cents. 

Hence  Division  may  be  defined  — 

I.  The  method  of  finding  how  many  times  one  number  is 
contained  in  another  of  the  same  kind.     (Ex.  1.)     Or, 

II.  The  method  of  dividing  a  number  into  a  certain  num 
ber  of  equal  parts.     (Ex.  2.) 

The  Dividend  is  the  number  to  be  divided,  and  answers  to 
the  product  in  multiplication. 

The  Divisor  is  the  number  by  which  we  divide,  and  answers 
to  one  of  the  factors. 

The  Quotient  is  the  result  or  answer,  and  is  the  other  fac 
tor.  When  anything  is  left,  it  is  called  the  Remainder. 

NOTE.  — In  the  first  use  of  division,  the  divisor  and  dividend  must 
be  of  the  same  kind,  for  it  would  be  absurd  to  ask  how  many  times  a 
number  of  pounds  of  butter  is  contained  in  a  number  of  gallons  of 
molasses.  In  the  second  use,  the  quotient  is  of  the  same  kind  with 
the  dividend,  for  if  a  number  of  acres  of  land  should  be  divided  into 
several  parts,  each  part  will  still  be  acres  of  land. 

SIGN.  —  The  sign  of  division  is  a  short  horizontal  line  be 
tween  two  dots,  thus  -K  This  shows  that  the  number  before 
it  is  to  be  divided  by  the  number  after  it ;  thus  27  -7-  9  =  3, 
is  read,  27  divided  by  9  is  equal  to  3 ;  or,  to  shorten  the  ex 
pression,  9  in  27  3  times.  Or  the  dividend  may  be  written 
in  place  of  the  upper  dot,  and  the  divisor  in  place  of  the  lower 
dot ;  thus  ^7-  shows  that  27  is  to  be  divided  by  &  as  before. 

Questions.  —  If  31.  In  what  way  is  the  first  example  performed? 
How  might  the  operation  be  shortened?  How  often  is  one  number  con 
tained  in  \nother?  What,  then,  is  division?  Show  its  relation  to  mul 
tiplication.  What  is  the  object  in  the  first,  and  what  in  the  second, 
example?  Define  division.  What  is  the  dividend?  to  what  does  it 
answer  in  subtraction,  and  to  what  in  multiplication  ?  What  the  divisor, 
and  to  what  does  it  answer  in  subtraction  and  multiplication  ?  What 
the  quotient,  and  to  what  does  it  answer  ?  Explain  the  divisor  and  divi 
dend  in  the  first  use  of  division.  The  dividend  and  quotient  in  the 
second. 

6 


50 


DIVISION    OF  SIMPLE   NUMBERS. 
DIVISIOX    TABLE. 


NOTE.  —  The  expression  used  by  the  pupil  in  reciting  the  table 
may  be,  2  in  2  one  time,  2  in  4  two  times,  4  in  12  three  times,  &c. 

1=1 


I        0 

2~ 

46-  =  4 


%°-  =  4  i  ^  =  4 
V  =  5  \  4*  =  5 


-¥-  ~~  4 
4-  =  6 

jf-  =  9 


o 


-6/1  ==  7 
¥  =  8 


28- 

-  7,or-2T8-  = 

how 

many? 

49- 

-  7, 

or   ^ 

LQ.  =  how 

many? 

42- 

-  6,or4ga  = 

how 

many? 

32- 

-  4, 

or   -[ 

5f  =  how 

many? 

54- 

-  9,  orAf  = 

how 

many? 

99- 

-11, 

or   j 

^f  =  how 

many? 

32- 

-  8,or-3/  = 

how 

many  ? 

84- 

-12, 

or    ^ 

rf  =  how 

many? 

33- 

-ll,orff  == 

how 

many? 

108- 

-12, 

or-V 

^  =  how 

many? 

NOTE.  — The  pupil  should  be  thoroughly  exercised  in  the  foregoing 
table. 

^T  S3.  The  principles  of  division  will  be  made  more  plain 
to  the  pupil  by  turning  his  attention  to  the  same  diagram  to 
which  it  was  directed  while  illustrating  the  principles  of  mul 
tiplication,  since  division  is  the  reverse  of  multiplication. 

DIAGRAM  OF  STARS. 

In  multiplication,  we  call  the  whole  num- 
^     y?     ^     %  ber  of  stars  a  symbol  of  the  product. 

In  division,  a  symbol  of  the  dividend. 


1F  33-35.  DIVISION  OF  SIMPLE  NUMBERS.  51 

In  multiplication,  stars  in  a  row,  and  number  of  rows,  are 
symbols  of  the  multiplicand  and  multiplier, 
which  are  factors  of  the  product. 

In  division,  stars  in  a  row,  and  rows  of  stars,  are  symbols  of 
the  divisor  and  quotient,  which  are  factors  of 
the  dividend. 

IT  33.  When  the  object  in  division  is  to  find  how  many 
times  one  number,  or  quantity,  is  contained  in  another  num 
ber,  or  quantity,  the  divisor  must  be  of  the  same  kind  as  the 
dividend,  (stars  in  a  row,)  and  the  quotient  will  be  a  number 
telling  how  many  times  (rows  of  stars.) 

On  the  other  hand — when  the  object  is  to  divide  a  number 
or  quantity,  into  a  given  number  of  equal  parts,  the  quotiem 
will  be  of  the  same  name  or  kind  as  the  dividend,  (stars  in  a 
row.)  If  we  divide  35  apples  into  5  parts,  the  quotient,  7 
apples,  will  be  one  part,  (stars  in  a  row,)  and  the  divisor,  5, 
will  be  a  number,  that  is,  the  number  of  parts,  (rows  of  stars.) 

IT  34.  It  has  been  remarked  that  division  is  a  short  way 
of  performing  many  subtractions.  How  often  car*  3  be  sub 
tracted  from  963  ?  Ans.  321  times.  To  set  down  963  and 
subtract  3  from  it  321  times  would  be  a  long  and  tedious  pro 
cess  ;  but  by  division  we  may  decompose  the  number  963 
thus?;  963  =  900  +  60  +  3,  and  say  3  is  contained  in  9(00,) 
3(013)  times, in  6(0,)  2(0)  times,  and  in  3,  1  time  =  321  times, 
which  brings  us  to  the  same  result  in  a  much  shorter  way. 

IT  3*1.  1.  How  many  yards  of  cloth,  at  3  dollars  a  yard, 
can  be  bought  for  936  dollars  ? 

SOLUTION.  —  As  many  yards  as  3  dollars  are  contained  times  in  936 

Questions.  — «[[  32.  What,  in  the  diagram  of  stars,  may  be  taken 
as  a  symbol  of  the  dividend  in  division  ?  Stars  in  a  row  and  ron-s  of  ftars 
are  taken  as  symbols  of  what,  in  multiplication  ?  of  what  in  division  ? 
If  the  dividend  be  103,  the  divisor  12,  and  the  quotient  9,  how  would  you 
make  a  diagram  to  correspond? 

^[  33.  When  the  object  is  to  find  how  many  times  one  number  or 
quantity  is  contained  in  another,  the  divisor  will  be  of  what  name  or 
kind  ?  the  quotient  will  be  what  ?  When  the  object  is  to  divide  a  num 
ber  or  quantity  into  a  given  number  of  parts,  the  quotient  will  be  of 
what  name  or  kind?  the  divisor  will  be  what? 

If  34.  How  can  you  make  it  appear  from  the  diagram  tha.t  division 
is  a  shorter  way  of  performing  many  si  .tractions?  Find  on  the  black 
board  the  quotient  of  963  divided  by  3  How  would  this  example  be 
performed  by  subtraction  ? 


52  DIVISION   OF  SIMPLE  NUMBERS.  H  35 

dollars,  or  as  many  as  3  can  be  subtracted  times  from  936  ;  936  dol 
lars  is  the  dividend,  (number  of  stars,)  3  dollars  (stars  in  a  row)  the 
divisor. 

PREPARATION.  Write  the  divisor  on  the  left  of  the  divi- 

Dividend,    dend,  separate  them  by  a  curved   line,  and 
Divisor,  3  )  936      '     draw  a  line  underneath. 

OPERATION.  We  may  decompose  the  dividend  thus,  936  = 

3  )  936     90°  4-30-4-6,  and  divide  each  part  separately. 

Beginning  at  the  left  hand,  we  say,  3  in  9,  3 
Quotient,  312    times.     This  quotient  3  is  3  hundred,  because  the 
9  which  we  divided  is  hundreds ;  therefore  we  write 
it  under  the  9  in  the  place  of  hundreds. 

Proceeding  to  the  next  figure,  we  say,  3  in  3,  1  time,  which,  being 
1  ten,  we  write  it  in  tens'  place.  Lastly,  3  in  6,  2  times,  which,  be 
ing  units,  we  write  the  2  in  units'  place,  and  the  work  is  done.  The 
quotient  (number  of  rows)  is  3  hundred,  (300,)  1  ten,  (10,)  and  2 
units,  or  312  yards,  Ans. 

NOTE.  —  The  quotient  figure  will  always  be  of  the  same  order  of 
units  as  the  figure  divided  to  obtain  it. 

2.  2846 -s-  2  =  how  many?   840 -f- 4  =  how  many  ?   500 
-5-  5  =  how  many  ? 

3.  If  you  give  856  dollars  to  4  men,  how  many  dollars  will 
you  give  to  each  ? 

OPERATION.  SOLUTION.  — Write  down  the  numbers 

Divisor,  Dividend,  as  before.   Divide  the  first  figure,  8,  (him- 

4  men, )  856  dollars,     dreds,)  in  the  dividend  as  before.     Pro 
ceeding  to  the  next  figure,  5,  (tens,)  4  is 

Quotient,  214  dollars,     contained  1  (ten)  time  in  4  of  the  tens, 
or  40,  and  there  is  1  ten  left,  which,  added 

to  the  6  units,  will  make  16,  and  4  in  16  units,  4  (units)  times.  Ans. 
214  dollars. 

Here  again  we  see  that  the  856  is  taken  in  three  parts,  800,  40, 
and  16,  and  each  part  is  divided  separately.  When  this  decomposing 
into  parts  can  be  done  in  the  mind,  as  in  these  examples,  the  process 
is  called  Short  Division.  It  can  always  be  done  when  the  divisor 
does  not  exceed  12. 

4.  Answer  the  following  questions  after  the  same  manner, 
viz.,  650  -r-5=  how  many  ?     8490  -i-  6  ;  or,  what  expresses 
the  same  thing,  -ft^ao.  _  how  many  ?     &i|Aa  =  how  many  ? 
8.QQ2&  __  jlow  many  ? 

5.  What  is  the  quotient  of  14371  divided  by  7  ? 

OPERATION.  Ther:  are  two  other  things  to  be  learned  in 

7)14371  this  opt  ration.     First,  the  divisor,  7,  is  net  con- 

_          tained   a  1,  the  first  figure  of  the  dividend     then 

Quotient^     2053  take  two  figures,  or  so  many  as  shall  contain  the 


1"  35.  DIVISION  OF   SIMPLE  NUMBERS.  5$ 

divisor,  and  say,  7  in  14,  2  times;    we  write  2  in  the  quotient,  in 
thousands'  place,  because  we  divided  14  thousands. 

Then,  again,  proceeding  to  the  next  figure,  3  in  the  dividend  will  not 
contain  the  divisor,  7 ;  to  obviate  this  difficulty,  we  place  a  cipher  in 
the  quotient,  joining  the  3  to  the  7  tens,  calling  it  37  tens,  and  so  pro- 
coed.  Ans.  2053. 

Hence,  for  Short  Division,  this  general 

RULE. 

I.  Write  the  divisor  at  the  left  hand  of  the  dividend ,  sep 
arate  them  by  a  line,  and  draw  a  line  under  the  dividend,  to 
separate  it  from  the  quotient. 

II.  Find  how  many  times  the  divisor  is  contained  in  the 
first  left  hand  figure  or  figures  of  the  dividend,  and  place  the 
result  directly  under  the  last  figure  of  the  dividend  taken,  for 
the  first  figure  of  tjie  quotient. 

III.  If  there  be  no  remainder,  divide  the  next  figure  in  the 
dividend  in  the  same  way;  but,  if  there  be  a  remainder,  join 
it  to  the  next  figure  of  the  dividend  as  so  many  tens,  and  then 
find  how  many  times  the  divisor  is  contained  in  this  amount, 
and  set  down  the  result  as  before. 

IV.  Proceed  in  this  manner  till  all  the  figures  in  the  divi 
dend  are  divided. 

EXAMPLES    FOR    PRACTICE. 

6.  A  man   has  256  hours'  work  to  do ;  how  many  days 
will  it  take  him,  if  he  work  8  hours  each  day  ? 

"Ans.  32  days. 

7.  **^^==  haw  many  1  Ans.  215477. 

8.  In  1  gallon  are  4  quarts  ;   how  many  gallons  in  2784 
quarts  ?  Ans.  696  gallons. 

9.  Seven  men  undertake  to  build  a  barn,  for  which  they 
are  to  receive  602  dollars ;  into  how  many  equal  parts  must 

the  money  be  divided  ?     How  much  will  1  part  be  ?  3 

parts  ?  -     -  5  parts  ?  (See  If  30.) 

Ans.  to  the  last,  430  dollars. 


Questions.  —  ^[  35.  When  the  dividend  is  large,  how  must  it  be 
taken  ?  ho\v  divided  ?  How  is  it  done  when  the  divisor  does  not  exceed 
12  ?  What  is  the  preparation  ?  Where  do  you  begin  the  division  ?  If 
you  divide  units,  what  will  the  quotient  be  ?  if  tens,  what  ?  hundreds, 
what?  If  at  any  time  you  have  a  remainder,  what  do  you  do  with  it? 
In  Ex.  5.  there  are  two  things  to  be  learned;  what  is  the  first  thing? 
the  second  thing  ?  What  then  is  to  be  done  ?  How  do  you  obviate  this 
difficulty?  What  does  the  cipher  you  write  in  the  quotient  show? 
What  is  short  division  ?  When  employed  ?  Repeat  the  rule 

5* 


54  DIVISION  OF  SIMPLE  NUMBERS.  1T  36 

10.    Divide  24108  by  ]  2.  Quotient,  2009. 

IF  SO.  1.  A.  man  gave  86  apples  to  5  boys ;  how  many 
apples  did  each  boy  receive  ? 

Dividend,  SOLUTION.  —  Here,    dividing 

Divisor,  5 )  86  the  number  of  the  apples  (86) 

by  the  number  of  boys,  (5,)  we 
Quotient,  171  Remainder.  find  that  each  boy's  share  would 

be  17  apples ;  but  there  is  1  ap 
ple  left,  and  this  apple,  which  is  called  the  remainder,  is  a  portion  of 
the  dividend  yet  undivided.  Wherefore  this  1  apple  must  be  divided 
equally  among  the  5  boys.  But  when  a  thing  is  divided  into  5  equal 
parts,  one  of  the  parts  is  called  .*.,  (^[  30.)  So  each  boy  will  have  -J-  of 
an  apple  more,  or  17-^  apples  in  all.  Ans.  17^-  apples. 

•    NOTE  1. — The  17  (apples)  expressing  whole  apples,  are  called 
Integers,  that  is,  whole  numbers1. 

Integers  are  numbers  expressing  w/iole  things ;  thus,  86  oranges, 
4  dollars,  5  days,  75,  268,  &c.,  are  integers,  or  whole  numbers. 

NOTE  2. — The  £  (1  fifth)  of  an  apple  given  to  each  boy,  ex 
pressing  part  of  a  divided  apple,  is  called  a  Fraction,  or  broken  num 
ber. 

Fractions  are  the  parts  into  which  a  unit  or  whole  thing  may  be 
divided.  Thus,  £  (1  half)  of  an  apple,  §  (2  thirds)  of  an  orange,  % 
(4  sevenths)  of  a  week,  are  fractions. 

NOTE  3.  —  A  number  composed  of  a  whole  number  and  a  fraction, 
is  called  a  Mixed  Number ;  thus,  the  number  17-£  (apples)  in  the 
above  example,  is  a  mixed  number,  being  composed  of  the  integers  17 
and  the  fraction  •£. 

If  we  examine  the  fraction,  we  shall  see,  that  it  consists  of  the  re 
mainder  (1)  for  its  numerator,  and  the  divisor  (5)  for  its  denominator. 
Therefore,— 

If  there  be  a  remainder,  set  it  down  at  the  right  hand  of  the  quo 
tient  for  the  numerator  of  a  fraction,  under  which  write  the  divisor  for 
its  denominator. 

2.  Eight  men  drew  a  prize  of  453  dollars  in  a  lottery; 
how  many  dollars  did  each  receive  ? 

Dividend,        Here,  after  carrying  the  division  as  far  as 
Divisor,  8 )  4-53  possible   by  whole  numbers,  we  have  a  re 

mainder  of  5  dollars,  which,  written  as  above 

Quotient,        56|          directed,  gives  for  the  answer  56  dollars  and  $ 
(5  eighths)  of  another  dollar,  to  each  man. 

Questions.  —  IT  36.  What  are  integers ?  fractions?  a  mixed  num 
ber?  If  ihere  be  a  remainder  after  division,  it  is  a  portion  of  what? 
What  do  ;ou  do  with  it?  If  you  have  a  quotient  of  23T9T,  what  was 
the  rem«u  «der  ?  What  was  rt  e  divisor  ? 


1T37.  DIVISION  OP  SIMPLE  NUMBERS.  55 

IT  37.     PROOF. 

1.  16  X  5  =  80,  Product.   )      Multiplication  and  division 

2.  Dividend,  80  -7-  5=  16.  \  are  the  reverse  of  each  other. 

We  see,  in  the  2d  of  the  above  examples,  that  the  product 
80  of  the  1st  example,  divided  by  5,  one  of  its  factors,  brings 
out  16,  the  other  factor,  and  hence  that  division  may  be  used 
to  prove  multiplication.  We  see,  also,  in  the  1st  example,  that 
the  divisor  and  quotient  of  the  2d  example,  multiplied  togeth 
er,  reproduce  the  dividend,  and  hence  that  multiplication  may 
be  used  to  prove  division.  Hence  the 

RULE. 

To  prove  multiplication  ly  To  prove  division  ly  mul- 
division. —  Divide  the  prod-  tiplication. — Multiply  the  di- 
uct  by  one  factor,  and,  if  the  visor  and  quotient  together, 
work  be  right,  the  quotient  and  if  the  work  be  right,  the 
will  be  the  other  factor.  product  will  be  equal  to  the 

dividend. 

NOTE  1.  —  To  prove  division,  if  there  be  a  remainder.  Multiply  the 
integers  of  the  quotient  by  the  divisor,  and  to  the  product  add  the  re 
mainder.  If  the  work  be  right,  their  sum  will  be  equal  to  the  divi 
dend. 

Example. — Divide  1145  by  7. 

OPERATION.  PROOF. 

7 )  1145  163  integers  of  the  quotient. 

7  divisor. 

163f  

1141 

4  remainder  added. 

1145  =  the  dividend. 

NOTE  2 .  —  Proof  ly  excess  of  nines.  Find  the  excess  of  nines  in  the 
divisor,  write  it  before  the  sign  of  multiplication,  also  in  the  quotient, 
and  write  it  after  the  sign  ;  multiply  together  these  excesses,  and 
write  the  excess  of  nines  in  their  product  over  the  sign  ;  subtract  the 
remainder,  if  any,  from  the  dividend,  and  write  the  excess  of  nines  in 
what  is  left  under  the  sign.  If  the  numbers  under  and  over  the  sign 
be  alike,  the  work  is  presumed  to  be  right,  in  accordance  with  princi 
ples  explained  in  multiplication,  ^[  23,  note  3. 

Questions.  —  IT  37.  To  what,  in  multiplication,  does  the  dividend 
in  division  answer?  To  what,  the  divisor  and  quotient?  How,  then, 
is  multiplication  proved  by  division?  How  division  by  multiplication? 
How,  when  there  is  a  remainder  ? 


56  DIVISION   OF  SIMPL!  i  NUMBERS.  1  38 

Let  the  pupil  be  required  to  prove  tl  e  examples  which  follow. 
EXAMPLES    FOR    PRACTICE. 

1.  Divide  1005903360  by  2,  3,  4,  5,  6,  7,  8,  9,  10,  11 

and  12. 

2.  If  2  pints  make  a  quart,  Low  many  quarts  in  8  pints  ? 

-  in  12  pints  ?  -  in  20  pints  ?  -  in  24  pints  ?  - 
in  248  pints  ?  -  in  3764  pints  ?  -  in  47632  pints  ? 

Ans.  to  the  last,  23816  quarts. 

3.  Four  quarts  make  a  gallon  ;  how  many  gallons  in  8 
quarts  ?  -  in  12  quarts  ?   -  in  20  quarts  ?   -  in  36 
quarts  ?  -  in  368  quarts  ?  -  in  4896  quarts  ?  -  in 
5436144  quarts  ?  Ans.  to  the  last,  1359036  gallons. 

4.  There  are  7  days  in  a  week  ;  how  many  weeks  in  365 
days  ?  Ans.  52j-  weeks. 

5.  When  flour  is  worth  6  dollars  a  barrel,  how  many  bar 
rels  may  be  bought  for  25  dollars  ?  how  many  for  50  dollars? 

-  for  487  dollars  ?  -  for  7631  dollars  ? 

6.  Divide  640  dollars  among  4  men. 

640  -J-  4,  or  £££  =  160  dollars,  Ans. 

7.  678-7-6,  or  &%8-=  how  many?  Ans.  113. 

8.  AOA4  =  how  many?  Ans.  1008. 

9.  i^t==  how  many?  Ans.  1033f 

10.  -HP  __  how  many  ? 

11.  £ff^=  how  many  ? 

12.  AAJJLI  —  how  many  ? 
13. 


^F  38.     1.    Divide  4478  dollars  equally  among  21  men. 

SOLUTION.  —  When,  as  in  this  example,  the  divisor  exceeds  12, 
the  decomposing  into  parts  cannot  be  done  in  the  mind  as  in  short  di 
vision,  but  the  whole  process  must  be  written  down  at  length  in  the 
following  manner. 

OPERATION.  We  say>  21  in  44'   (hundreds,)  2 

(hundred)  times,  and  write  2  on  the 

Dvfr.    Divd.    Quot.  right  hand  of  the  dividend  for  the  first 

21  )  4478  (  213^j-  figure  of  the  quotient.     That  is,  we 

42   1st  part.  have  2  hundred  dollars  for  each  of  21 

men,  requiring  21  X  2  (hundred)  =42 

hundred  in  all.     This  is  the  first  part 

21J2dpart.  divided.     The  42  hundred  must  now 

~7o  be  subtracted  from  the  hundreds  in  the 

^  dividend,  and  we  find  2  (hundred)  re- 

3  3(2  part.  maining,  to  which,  bringing  down  the 

~K    T>»™~^J*~      7  tens,  the  whole  is  27  tens.     21  in 
5  Remainder     g?j          }  l  (ten)  ^    Each 


1T38.  DIVISION  OP  SIMPLE  NUMBERS.  57 

has  now  10  dollars  more,  which  require  21  tens,  the  second  part,  and 
taking  this  from  the  27  tens,  and  bringing  down  the  8  units,  we  have 
68  dollars  yet  to'  be  divided.  21  in  68,  3  times,  that  is,  each  man 
will  have  3  dollars,  which  will  require  63  dollars,  the  third  part,  and 
there  are  5  dollars  left.  This  will  not  give  each  man  a  whole  dollar, 
but  2=\-  of  a  dollar.  So  each  man  has  2  hundred,  1  ten,  3^j-  dollars  ; 
that  is,  213^  dollars,  Ans. 

The  parts  into  which  the  dividend 
is  decomposed,  are  42  hundreds,  which 

\st  part,  4200  dollars,  contain  the  divisor  2  (hundred)  times  ; 
^d  part,  210  "  2i  tens>  which  contain  the  divisor  1 
3d  Tjart  63  "  (ten)  time ;  and  63,  which  contain 

r>         •  ' j  c  the   divisor  3   (units)   times,    or   213 

Remainder, 5  timeg  ^  ^  ^   ^  rem;inder  5> 

4478        "          We   here  see   that  the   parts   added 
make  the  whole  sum. 

This  method  of  performing  the  operation  is  called  Long 
Division.  It  consists  in  writing  down  the  whole  work  of  di 
Tiding,  multiplying,  and  subtracting. 

From  the  illustrations  now  given,  we  deduce  the  following 

RULE. 

To  perform  Long  Division. 

I.  Place  the  divisor  at  the  left  hand  of  the  dividend,  and 
separate  them  by  a  curved  line,  and  draw  another  curved  line 
on  the  right  of  the  dividend,  to  separate  it  from  the  quotient. 

II.  Take  as  many  figures  on  the   left  of  the   dividend  as 
will  contain  the  divisor  one  or  more  times ;  find  how  many 
times  they  contain  it,  and  put  the  answer  at  the  right  hand  of 
the  dividend  for  the  first  figure  in  the  quotient. 

III.  Multiply  the  divisor  by  this  quotient  figure,  and  set  the 
product  under  that  part  of  the  dividend  which  you  divided. 

IV.  Subtract  this  product  from  the  figures  over  it,  and  to 
the  remainder  bring  down  the  next  figure  in  the  dividend. 

V.  Divide  the  number  this  makes  up  as  before.    Continue 
to  bring  down  and  divide  until  all  the  figures  in  the  dividend 
have  been  brought  down  and  divided. 

PROOF.  —  Long  division  may  be  proved  by  multiplication, 
by  the  excess  of  nines,  by  adding  up  the  parts  into  which  the 

Questions.  —  1[  38.  What  cannot  be  done,  when  the  divisor  ex 
ceeds  12  ?  Into  what  parts  is  the  dividend,  in  the  first  example,  decom 
posed  ?  How,  and  for  what,  is  42  obtained  ?  21?  63?  Explain  the 
proof.  What  is  long  division,  and  in  what  does  it  consist  ?  Give  the 
rule.  Name  the  different  methods  of  proof.  Give  the  substance  of  note 
1 ;  note  2  ;  note  3. 


68  DIVISION  OF  SIMPLE  NUMBERS.  If  38 

dividend  is  decomposed,  or  by  subtracting  the  remainder  from 
the  dividend  and  dividing  what  is  left  by  the  quotient,  which 
if  the  work  is  right,  will  bring  the  divisor. 

NOTE  1.  —  Having  brought  down  a  figure  to  the  remainder,  if  the 
number  it  makes  up  will  not  contain  the  divisor,  write  a  cipher  in  the 
quotient,  and  bring  down  the  next  figure. 

NOTE  2.  —  When  we  multiply  the  divisor  by  any  quotient  figure, 
and  the  product  is  greater  than  the  number  we  divided,  the  quotient 
figure  is  too  large,  and  must  be  diminished. 

NOTE  3. — If  the  remainder,  at  any  time,  be  greater  than  the  divi 
sor,  or  equal  to  it,  the  quotient  figure  is  too  small,  and  must  be  in 
creased. 

EXAMPLES  FOR  PRACTICE. 

1.  How  many  hogsheads  of  molasses,  at  27  dollars  a  hogs 
head,  may  be  bought  for  6318  dollars  ? 

Ans.  234  hogsheads. 

2.  If  a  man's  income  be  1248  dollars  a  year,  how  much  is 
that  per  week,  there  being  52  weeks  in  a  year  ? 

Ans.  24  dollars  per  week. 

3.  What  will  be  the  quotient  of  153598,  divided  by  29  ? 

Ans.  5296£f . 

4.  How  many  times  is  63  contained  in  30131  ? 

Ans.  478££  times ;  that  is,  478  times,  and  £f  of  another 
time. 

5.  What  will  be  the  several  quotients  of  7652,  divided  by 
16,  23,  34,  86,  and  92  ?  Ans.  to  the  last,  83^f . 

6.  If  a  farm,  containing  256  acres,  be  worth  7168  dollars, 
what  is  that  per  acre  ?  Ans.  28  dollars. 

7.  What  will  be  the  quotient  of  974932,  divided  by  365? 

Ans.  2671^-. 

8.  Divide  3228242  dollars  equally  among  563  men  ;  how 
many  dollars  must  each  man  receive  ?      Ans.  5734  dollars. 

9.  If  57624  be  divided  into  216,  586,  and  976  equal  parts, 
what  will  be  the  magnitude  of  one  of  each  of  these  equal 
parts  ? 

Ans.  The  magnitude  of  one  of  the  last  of  these  equal  parts 
will  be  59/TV 

10.  How  many  times  does  1030603615  contain  3215? 

Ans.  320561  times. 

11.  The  earth,  in  its  annual  revolution  round  the  sun,  is 
said  to  travel  596088000  miles ;  what  is  that  per  hour,  there 
being  8766  hours  in  a  year  ?  Ans.  68000  miles. 

12.  -i^34| gfAAa  =  how  many ?  Ans.  944581-^- 

13.  ^o|o 2- *•  ==  how  many  ?  Ans. 

14.  ^64i=nowman?  Ans. 


1F  39,  40.  DIVISION  OF  SIMPLE  NUMBERS.  59 

IT  4O.    Contractions  in  Division. 

IT  39.     I.    When  the  divisor  is  a  composite  number. 

1.  Bought  18  yards  of  cloth  for  72  dollars ;  how  much  was. 
that  a  yard  ? 

SOLUTION.  —  This  example  is  the  reverse  of  Ex.  1,  ^f  24.  It  was 
there  shown  that  3  and  6  are  factors  of  18  (3X6=  18.) 

If  the  18  yards  be  di- 

OPERATION.  Vided  into  3  pieces,  then 

3 )  72  dollars,  cost  of  18  yards.  the  cost  of  1  piece  would 

6 )  24  dollars,  cost  1  piece  =  6  yards,     be  one  third  as  much  as 

the  cost  of  3  pieces,  that 

Ans.  4  dollars,  cost  of  1  yard.  is,  72  -^-  3  =  24  dollars ; 

and  the  cost  or  1   yard 

would  be  one  sixth  of  the  cost  of  6  yards,  that  is,  24-7-6  =  4  dollars. 
That  is,  we  divide  the  price  of  18  yards  by  3,  and  get  the  price  of  one 
Vhird  of  18,  or  6  yards,  and  divide  the  price  of  6  yards  by  6,  and  get 
Jie  price  of  1  yard.  Ans.  4  dollars. 

Hence,  To  perform  division  when  the  divisor  is  a  composite 
number, 

RULE. 

1.  Divide  the  dividend  by  one  of  the  component  parts,  and 
the  quotient  arising  from  that  division,  by  the  other. 

II.  If  the  component  parts  be  MORE  than  two.  —  Divide  by 
each  of  them  in  order,  and  the  last  quotient  will  be  the  quo 
tient  required. 

EXAMPLES    FOR    PRACTICE. 

2.  If  a  man  travel  28  miles  a  day,  how  many  days  will  it 
take  him  to  travel  308  miles  ?    4  X  7  =  28.    Ans.  1 1  days. 

3.  Divide  576  bushels  of  wheat  equally  among  48  (8  X  6) 
men.  Ans.  12  bushels  each. 

4.  Divide  1260  by  63  (=7  X  9)         Quotient  20.  Ans. 

5.  Divide  2430  by  81  (=  )  "       30.  A?ts. 

6.  Divide  448  by  56  (=  )  "         8.  Ans. 

IF  4O.  It  not  unfrequently  happens  that  there  are  remain 
ders  after  the  several  divisions,  as  in  the  following  example. 

1.  A  man  wished  to  carry  783  bushels  of  wheat  to  market , 
how  many  loads  would  he  have,  allowing  36  bushels  to  a 
load? 

Questions.  —  ^[39.  How  may  you  contract  the  operation  in  di 
vision  when  the  divisor  is  a  composite  number  ?  Which  factor  should 
you  divide  by  first  ?  Repeat  the 


6*0  DIVISION   OF  SIMPLE   NUMBERS.  If  40. 

SOLUTION.  —  First,  suppose  his  wheat  put  into  barrels,  each  barrel 
containing-  4  bushels.  It  would  take  as  many  barrels  as  4  is  contained 
times  in  783. 

It  would  take  195  barrels,  and  leave  a  remainder 
Z  _  ,          of  3  bushels. 
19o  3rem. 

Next,  suppose  he  takes  9  barrels  at  each  load ;  9  (barrels)  X  4 
(the  number  of  bushels  in  each  barrel)  =  36  bushels  at  a  load,  and  he 
would  have  as  many  loads  as  the  number  of  times  9  barrels  are  con 
tained  in  195  barrels. 

9)  195  Hence  we  see,  that  he  would  have  21  loads,  and 

leave  a  remainder  of  6  barrels ;    also,  a  former  re- 
21   6rem.    mainder  of  3  bushels. 

f  4)783  bushels. 

The  whole         

operation     ^  9)195  barrels,  and  3  bushels  remainder. 

stands  thus  :        

L       21  loads,  and  6  barrels  remainder. 

Onr  object  now  is  to  find  the  true  remainder.  The  last  remainder, 
6  barrels,  multiplied  by  the  first  divisor,  4,  which  is  the  number  of 
bushels  in  a  barrel,  gives  a  product  of  24  bushels.  To  this  add  the 
first  remainder,  3  bushels,  and  we  have  the  true  remainder,  27  bushels. 

Therefore,  When  there  are  remainders  in  dividing  by  TWO 
component  parts  of  a  number,  lo  get  the  TRUE  remainder, 

RULE. 

1.  Multiply  the  last  remainder  by  the  first  divisor,  and  to 
the  product  add  the >  first  remainder ;  the  sura  will  be  the  true 
remainder. 

II.  When  there  are  MORE  than  TWO  'itisors. — Multiply 
each  remainder,  except  that  from  the  first  divisor,  by  all  the 
divisors  preceding  the  divisor  which  gave  it ;  to  the  sum  of 
their  products  add  the  remainder  from  'ne  first  divisor,  if  any. 
and  the  amount  will  be  the  true  rema.v*der. 

2.  5783  •+•  108  =  how  many  ?     f  x  4  X  9=  108;  hence, 
three  divisors. 


Questions.—  If  4O.  When  there  are  remainders  in  dividing  by 
two  component  parts,  how  do  you  find  .he  true  remainder  ?  When  there 
are  more  than  two.  how?  If  there  '«  a  remainder  by  the  first  divisoi 
only  waut  is  the  true  remainder  ?  if  by  the  2d  divisor  and  none  by  the 
1st.  how  do  you  obtain  the  true  remainder  ?  if  by  the  3d,  and  none  by 
the  2d  and  Is;,  how?  if  by  the  l^t  and  3d,  and  "none  by  the  2d  how? 
Repeat  the  rule. 


1T41.  DIVISION  OF  SIMPLE  NUMBERS.  61 

OPERATION 

3)5783 

4)1927  and  2,  1st  rem. 


9 )    481  and  3,  2d  rem. 

53  and  4,  3d  rem. 

3d  rem.  4  X  4  (2d  div.)  X  3  (1st  div.)  =  48 
2d  rem.  3  X  3  (Irf  div.)  =    9 

1st  rem.  2  added    2 

True  rem.  59 
Ans.  S3$fg. 

NOTE.  — The  remainder  by  the  1st  divisor,  if  there  be  no  other,  is 
the  true  remainder. 

EXAMPLES  FOR  PRACTICE. 

3.  Divide  26406  by  42  =  6  X  7 ;   what  will  be  the  true 
remainder  ?  Ans.  30. 

4.  Divide  64823  by  3  component  parts,  the  continued  pro 
duct  of  which  is  98 ;  what  will  be  the  true  remainder  ? 

Ans.  23. 

5.  What  is  the  quotient  of  6811  divided  by  the  component 
parts  of  81  ?  Ans.  84BV 

6.  Divide  25431  by  the  component  parts  3  X  4  X  8  =  96, 
first,  in  the  order  here  given  ;  secondly,  in  a  reversed  order, 
8,  4,  3 ;  and  lastly,  in  the  order  4,  3,  8,  and  bring  out  the 
true  quotient  in  each  case.  Quotient,  264f£. 

If  41.     When  the  divisor  is  10,  100,  1000,  <£c. 

1.    A  prize  of  2478  dollars  is  drawn  by  10  men ;  what  is 
each  man's  share  ? 

OPERATION.  SOLUTION.  —  It  has  been  shown  (^[25)  that  an- 

10 )  2478  nexing  a  cipher  to  any  number  is  the  same  as  mul- 

tiplying  it  by  10  ;  the  reverse  of  this  is  equally  true  • 

Ozf"  8          ^  we  cut  °^ tne  "?nt  nand  iigure  from  any  number 

**7>nr       it  is  the  same  as  dividing  it  by  10  ;  the  figures  at  the 

left  will  be  the  quotient.     The  figure  8,  at  the  right, 

Shorter  way.     being  an  undivided  part,  is  the  remainder,  and  may 

QUOT.  REM.        be  written  over  the  divisor  («jj  36)  thus,  TV     We 

247  j  8          see  that  7,  which  was  tens  before,  is  made  units  ;  4, 

which  was  hundreds,   is  tens,  &c.     On  the  same 

principle,  if  we  cut  off  two  figures  it  is  the  same  as  dividing  by  100  ; 

it' three  figures,  the  same  as  dividing  by  1000,  &e. 

6 


62  DIVISION  OF  SIMPLE  NUMBERS.  IT  42 

Hence,  To  divide  by  1  with  any  number  of  ciphers  annexed, 

RULE. 

Cut  off,  by  a  line,  as  many  figures  from  the  right  hand  of 
the  dividend  as  there  are  ciphers  in  the  divisor. 

The  figures  at  the  left  of  the  line  will  be  the  quotient,  and 
those  at  the  right  the  remainder. 

EXAMPLES    FOR    PRACTICE. 

2.  A  manufacturer  bought  42604  pounds  of  wool  in  100 
days  :  l^ow  many  pounds  did  he  average  each  day? 

42604  -r-  100  =  426^,  or  426 1 04  ==  426T^  pounds,  Am. 

3.  In  one  dollar  are  100  cents  ;  how  many  dollars  in  42425 
cents  ?  Ans.  424-£&  ;  that  is,  424  dollars,  25  cents. 

4.  1000  mills  make  one  dollar  ;  how  many  dollars  in  4000 
mills?  in  25000  mills?  in  845000? 

Ans.  to  the  last,  845  dollars. 

5.  In  one  cent  are  10  mills  ;  how  many  cents  in  40  mills? 

in  400  mills  ?    in  20  mills  ? in  468  mills  ? 

in  4603  mills  ?  Ans.  to  the  last,  460-&  cents. 

IT  45*.  III.  When  there  are  ciphers  on  the  right  hand  of 
the  divisor. 

1.  A  general  divided  a  prize  of  749346  dollars  equally 
among  an  army  of  8000  men;,  what  did  each  receive  ? 

8 1  000 )  749 1 346  SOLUTION  .  —  The  divisor  8000 

.  is  a  composite  number,  of  which 

93  5  rem.  8  an(*  100°  are  component  parts. 

Dividing-  what  8000  men  receive 

_   ,rt  ,  .  innrk  r-r\r\(\          DY   1000,   which   VVC  do    by  CUttl!!? 

5  (2<Z  rem.}  X  1000  =  5000,     0£the  three  right  hand  fag  of 
ana  5000  +  346  (1st  rem.}  =     the  dividend,  we  get  749  dollar,*,, 
5346,  true  remainder.  which  8  men  will  receive,  with  a 

remainder  of  346  dollars  ;  and  di 
viding  749  dollars,  which  8  men  receive,  by  8,  we  get  what  1  man 
receives,  which  is  93  dollars,  and  a  remainder  of  5.  The  5  must  b<5 
multiplied  by  the  first  divisor,  1000,  and  the  first  remainder  added  to 
the  product ;  or,  which  is  the  same  thing,  (^f  25,)  the  first  remainder, 
346,  may  be  annexed  to  the  5,  and  we  have  the  Ans,.  93f-$^jy  dolls. 


Questions.  —  ^f  41.  If  we  annex  one  cipher  to  any  number,  how 
does  it  affect  it  ?  if  two  ciphers,  how  ?  three  ?  &c.  If  we  remove  the 
right  hand  figure  from  any  number,  what  is  the  result  ?  How  do  you 
divide  by  1  with  any  number  of  ciphers  annexed  ?  What  will  express 
the  remainder  ?  How  do  you  divide  by  10  ?  by  100  ?  by  1000  ?  by 
10000?  &c. 


1[43.  DIVISION  OF  SIMPLE  NUMBERS.  63 

Hence,  When  there  are  ciphers  on  the  right  hand  of  the 
divisor, 

RULE. 

1.  Cut  them  off,  and  also,  as  many  figures  from  the  right 
hand  of  the  dividend. 

II.  Divide  the  remaining  figures  in  the  dividend  by  the 
remaining  figures  in  the  divisor. 

III.  Annex  the  figures  cut  off  from  the  dividend  to  the 
remainder  for  the  true  remainder. 

EXAMPLES    FOR    PRACTICE. 

2.  In    1   square  mile  are  640  square  acres ;    how  many 
square  miles  in  23040  square  acres  ? 

Ans.  36  square  miles. 

3.  Divide  46720367  by  4200000.         Quot.  H^VAV 

4.  How  many  acres  of  land  can  be  bought  for  346500  dol 
lars,  at  20  dollars  per  acre  ?  Ans.  17325  acres. 

5.  Divide  76428400  by  900000.  Quot. 

6.  Divide  345006000  by  84000.  Quot.  4107£ 

7.  Divide  4680000  by  20,  200,  2000,  20000,  300,  4000, 
50,  600,  70000,  and  80.  Ans.  to  9th, 


IF  43.    Review  of  Division. 

Questions.  —  What  is  division?  In  what  does  the  process  consist? 
Define  it.  The  dividend  answers  to  what  in  subtraction  and  multiplica 
tion,  and  why  ?  the  divisor?  the  quotient?  In  what  ways  is  division 
expressed?  Apply  the  diagram  oi'  stars  to  division.  How  does  long 
division  differ  from  short  division?  Why  the  difference?  Rule  for 
short  division  —  for  long  division.  Give  the  different  methods  of  proof 
introduced  in  both.  To  what  does  a  remainder  give  rise,  and  how  writ 
ten?  What  are  fractions?  When  the  divisor  is  a  composite  number, 
how  do  you  proceed  ?  How  are  the  remainders  treated  ?  How  divide 
by  10,  100,  &c.  ?  How,  wnen  there  are  ciphers  at  the  right  hand  of  the 
divisor  ? 

EXERCISES. 

1.  An  army  of  1500  men,  having  plundered  a  city,  took 
2625000  dollars  ;  what  was  each  man's  share  ? 

Ans.  1750  dollars. 

2.  A  certain  number  of  men  were  concerned  in.  the  pay 
ment  of  18950  dollars,  and  each  man  paid  25  dollars  ;  what 
was  the  number  of  men?  Ans.  758. 

Questions.  —  ^[  42.  When  there  are  ciphers  on  the  right  hand  of 
the  divisor,  what  do  you  do  first  ?  How  do  you  divide  ?  How  do  you 
find  the  true  remainder?  Repeat  the  rule. 


64  DIVISION  OP  SIMPLE  NUMBERS.  f  43. 

3.  If  7412  eggs  be  packed  in  34  baskets,  how  many  in  a 
Nfatl  4?w.  218. 

4.  What  number  must  I  miutipiy  oy  135,  that  the  product 
may  be  505710  ?  Am.  3746. 

5.  Light  moves  with  such  amazing  rapidity,  as  to  pass 
from  the  sun  to  the  earth  in  about  8  minutes.     Admitting  the 
distance,   as  usually  computed,  to  be  95,000,000   miles,  at 
what  rate  per  minute  does  it  travel  ?    Am.  11875000  miles. 

6.  If  2760  men  can  dig  a  certain  canal  in  one  day,  how 
many  days  would  it  take  46  men  to  do  the  same?'   How 

many  men  would  it  take  to  do  the  work  in  15  days  ?  m 

5  days  ?    in  20  days  ?    in  40  days  ?   * in  120 

days  ? 

7.  If  a  carriage  wheel  turns  round  32870  times  in  running 
from  New  York  to  Philadelphia,  a  distance  of  95  miles,  how 
many  times  does  it  turn  in  running  1  mile  ?          Am.  346. 

8.  Sixty  seconds  make  one  minute ;   how  many  minutes 

in  3600  seconds?          -  in  86400  seconds?    in  604800 

seconds  ?   in  2419200  seconds  ? 

9.  Sixty  minutes  make  one  hour;    how  many  hours  in 
1440  minutes  ? in  10080  minutes  ?  in  40320  min 
utes  ?  in  525960  minutes  ? 

10.  Twenty-four  hours  make  a  day  ;   how  many  days  in 
168  hours  ?  in  672  hours  ?  in  8766  hours  ?    ' 

11.  How  many  times  can  I  subtract  forty-eisrht  from  four 
hundred  and  eighty!  Am.  10  times. 

12.  How  many  times  3478  is  equal  to  47854  ? 

Ans.  ISflfJ  times. 

13.  A  bushel  of  grain  is  32  quarts  ;  how  many  quarts  must 
I  dip  out  of  a  chest  of  grain  to  make  one  half  (£)  of  a  bushel  2 

for  one  fourth  (£)  of  a  bushel  ? for  one  eighth  (£) 

of  a  bushel  1  Ans.  to  the  last,  4  quarts. 

14.  Divide  9302688  by  648.  Quot.  14356. 

15.  Divide  1C30603615  by  3215.  Quot.  320561 

16.  Divide  5-221580  by  68705.  Quot.  76 

17.  Divide  2704503721  by  S3000. 

Quot.  33307,  rem.  22721. 

18.  If  the  dividend  be  275868665090130,  and  the  quotient 
562916859.  what  was  the  divisor?  Ans.  490070. 


1[44.  MISCELLANEOUS  EXERCISES.  65 

MISCELLANEOUS   EXERCISES, 

INVOLVING  THE  PRINCIPLES  OF  THE  PRECEDING  RULES. 

^T44.  The  four  preceding  rules,  viz.,  Addition,  Subtrac 
tion,  Multiplication,  and  Division,  are  called  the  Fundamental 
Rules  of  Arithmetic,  for  numbers  can  be  neither  increased  nor 
diminished  but  by  one  of  these  rules  ;  hence,  these  four  rules 
are  the  foundation  of  all  arithmetical  operations. 

EXERCISES    FOIl    THE    SLATE. 

1.  A  man  bought  a  chaise  for  218  dollars,  and  a  horse  for 
142  dollars ;  what  did  they  both  cost  ? 

2.  If  a  horse  and  chaise  cost  360  dollars,  and  the  chaise 
cost  218  dollars,  what  is  the  cost  of  the  horse  ? 

3.  If  the  horse  cost  142  dollars,  what  is  the  cost  of  the 
chaise  ? 

4.  If  the  sum  of  2  numbers  be  487,  and  the  greater  num 
ber  be  348,  what  is  the  less  number  ? 

5.  If  the  less  number  be  139,  what  is  the  greater  number? 

6.  If  the  minuend  be  7842,  and  the  subtrahend  3481,  what' 
is  the  remainder  ? 

7.  If  the  remainder  be  4361,  and  the  minuend  be  7842, 
what  is  the  subtrahend  ? 

8.  If  the  subtrahend  be  3431,  and  the  remainder  4361, 
what  is  the  minuend  ? 

9.  The  sum  of  two  numbers  is  48,  and  one  of  the  numbers 
is  19;  what  is  the  other? 

10.  The  greater  of  two  numbers  is  29,  and  their  difference 
10;  what  is  the  less  number? 

11.  The  less  of  two  numbers  is  19,  and  their  difference  k 
10 ;  what  is  the  greater  ? 

12.  The  sum  of  two  numbers  is  136,  their  difference  is  28, 
what  are  the  two  numbers  ?         ..       (  Greater  number,  8:2. 

s'  |  Less  number,       54. 

MENTAL   EXERCISES. 

1.  When  the  minuend  and  the  subtrahend  are  given,  how 
do  you  find  the  remainder  ?  Ex.  6. 

NOTE.  —  The  pupil  may  be  required  to  give  written  answers  to 
these  mental  exercises,  or  he  may  answer  orally;  in  either  case,  let 
6* 


66  MISCELLANEOUS  EXERCISES.  H  45. 

him  turn  to  the  exercise  for  the  slate  to  which  reference  is  made,  and 
let  him  apply  it  in  illustration  of  the  answer  he  gives.     Thus  — 

Ans.  Subtract  the  subtrahend  from  the  minuend,  and  the  differ 
ence  will  be  the  remainder,  as  Ex.  6,  (slate,)  where  the  minuend  and 
subtrahend  are  given  to  find  the  remainder,  —  we  subtract  the  subtra 
hend  3481  from  the  minuend  7842,  and  the  difference,  4361,  is  the 
remainder. 

2.  When  the  minuend  and  remainder  are  given,  how  do 
you  find  the  subtrahend  ?     Ex.  7. 

3.  When  the  subtrahend  and  the  remainder  are  given,  how 
do  you  find  the  minuend  ?     Ex.  8. 

4.  When  you  have  the  sum  of  two  numbers,  and  one  of 
them  given,  how  do  you  find  the  other  ?     Ex.  9. 

5.  When  you  have  the  greater  of  two  numbers,  and  their 
difference  given,  how  do  you  find  the  less  number  ?     Ex.  10. 

6.  When  you  have  the  less  of  two  numbers,  and  their  dif 
ference  given,  how  do  you  find  the  greater  number?    Ex.  11. 

7.  When  the  sum  and  difference  of  two  numbers  are  given, 
how  do  you  find  the  two  numbers  ?     Ex.  12. 

EXERCISES    FOR   THE    SLATE. 

If  45.  1.  If  the  multiplicand  (squares  in  a  row)  be  754, 
and  the  multiplier  (rows  of  squares)  be  25,  what  will  be  the 
product  (no.  of  squares)  ? 

2.  If  the  product  (no.  of  squares)  be  18850,  and  the  mul 
tiplicand   (squares  in  a  row)  be  754,  what  must  have  been 
the  multiplier  (rows  of  squares)  ? 

3.  If  the  product  (no.  of  squares)  be  18850,  and  the  mti.l- 
tiplier  (rows  of  squares)  be  25,  what  must  have  been  the 
multiplicand  (squares  in  a  row).? 

4.  If  the  dividend  (no.  of  squares)  be  144,  and  the  divi 
sor  (squares  in  a  row)  be  8,  what   is    the   quotient   (no.   of 
rows)  ? 

5.  If  the  dividend  (no.  of  squares)  be  144,  and  the  quo 
tient  (no.  of  rows)  be  18,  what  must  have  been  the  divisor 
(squares  in  a  row)  ? 

6.  If  the  divisor  (squares  in  a  row)  be  8,  and  the  quotient 
(rows  of  squares)  be  18,  what  must  have  been  the  dividend 
(no.  of  squares)  ? 

7.  The  product  of  three  numbers  is  52o,  and  two  of  the 
cumbers  are  5  arid  7,  what  is  the  other  number  ?   Ans.  15. 


IT  46.  MISCELLANEOUS  EXERCISES.  67 

MENTAL   EXERCISES. 

When  the  factors  are  given,  how  do  you  find  the  product  ? 
Ex.1. 

When  the  product  and  one  factor  are  given,  how  do  you 
find  the  other  ?  Ex.  2  and  3. 

When  the  dividend  and  quotient  are  given,  how  do  you 
find  the  divisor  ?  Ex.  5. 

When  the  divisor  and  quotient  are  given,  how  do  you  find 
the  dividend  ?  Ex.  6. 

When  the  product  of  three  numbers  and  two  of  them  are 
given,  how  do  you  find  the  other  ?  Ex.  7. 

EXERCISES    FOR    THE    SLATE. 

IT  46.  1.  What  will  be  the  cost  of  15  pounds  of  butter, 
it  13  cents  a  pound  ? 

2.  A  man  bought  15  pounds  of  butter  for  195  cents ;  what 
was  that  a  pound  ? 

3.  A  man  buying  butter,  at  15  cents  a  pound,  paid  out  195 
cents  ;  how  many  pounds  did  he  buy  ? 

4.  When  rye  is  75  cents  a  bushel,  what  will  be  the  cost  of 
084  bushels  ?  how  many  dollars  will  it  be  ? 

5.  If  984  bushels  of  rye  cost  738  dollars,  (73800  cents,) 
what  is  the  price  of  1  bushel  ? 

6.  A  man  boughi  rye  to  the  amount  of  738  dollars,  (73800 
tents,)  ai  75  cents  a  bushel ;  how  many  bushels  did  he  buy  ? 

7.  If  648  pounds  of  tea  cost  284  dollars,  (28400  cents,) 
*hat  is  the  price  of  1  pound  ?     28400  -r-  648=  how  many  ? 

MENTAL    EXERCISES. 

1.  When  the  price  of  one  pound,  one  bushel,  &c.,  of  any 
Commodity  is  given,  how  do  you  find  the  cost  of  any  number 
tf  pounds,  or  bushels,  &c.,  of  that  commodity?  Ex.  1  and  4. 
L"  the  price  of  the  1  pound,  &c.,  be  in  cents,  in  what  will  the 

\\  hole  cost  be  ?  if  in  dollars,  what  ?  if  in  shillings  ? 

if  in  pence  ?  &c. 

2.  When  the  cost  of  any  given  number  of  pounds,  or  bush 
els,  &c.,  is  given,  how  do  you  find  the  price  of  one  pound,  or 
bushel,  &c.  ?  Ex.  2,  5,  and  7.     In  what  kind  of  money  will 
the  answer  be  ? 

3.  When  the  cost  of  a  number  of  pounds,  &c.,  is  given,  and 
abo  the  price  of  one  pound,  &c.,  how  do  you  find  the  number 
of  pounds,  &c.  ?  Ex.  3  *>ad  6. 


68  MISCELLANEOUS    EXERCISES.  IT  47 

EXERCISES    FOR   THE    SLATE. 

^T4T.  1.  A  boy  bought  a  number  of  apples;  he  gave 
away  ten  of  them  to  his  companions,  and  afterwards  bought 
thirty-four  more,  and  divided  one  half  of  what  he  then  had 
among  four  companions,  who  received  8  apples  each ;  ho\v 
many  apples  did  the  boy  first  buy  ? 

Let  the  pupil  take  the  last  number  of  apples,  8,  and  reverse 
the  process.  Ans.  40  apples. 

2.  There  is  a  certain  number,  to  which  if  4  be  added,  and 
from  the  sum  7  be  subtracted,  and  the  difference  be  multiplied 
by  8,  and  the  product  divided  by  3,  the  quotient  will  be  64 ; 
what  is  that  number  ?  Ans.  27. 

3.  If  a  man  save  six  cents  a  day,  how  many  cents  would 

he  save  in  a  year,  (365  days  ?)  how  many  in  45  years  ? 

how  many  dollars  would  it  'be  ?  how  many  cows  could  he 
buy  with  the  money,  at  12  dollars  each  ? 

Ans.  to  the  last,  82  cows,  and  1  dollar  50  cents  remainder. 

4.  A  man  bought  a  farm  for  22464  dollars ;  he  sold  one 
half  of  it  for  12480  dollars,  at  the  rate  of  20  dollars  per  acre ; 
how  many  acres  did  he  buy  ?   and  what  did  it  cost  him  per 
acre  ?  Ans.  to  the  last,  18  dollars. 

5.  How  many  pounds  of  pork,  worth  6  cents  a  pound,  can 
be  bought  for  144  cents  ? 

6.  How  many  pounds  of  butter,  at  15  cents  per  pound,  must 
be  paid  for  25  pounds  of  tea,  at  42  cents  per  pound  ? 

7.  A  man  married  at  the  age  of  23 ;  he  lived  with  his  wife 
14  years ;  she  then  died,  leaving  him  a  daughter  12  years  of 
age ;  8  years  after,  the  daughter  was  married  to  a  man  5 
years  older  than  herself,  who  was  40  years  of  age  when  the 
father  died ;  how  old  was  the  father  at  his  death  ? 

Ans.  60  years. 

8.  The  earth,  in  moving  round  the  sun,  travels  at  the  rate 
of  68000  miles  an  hour ;  how  many  miles  does  it.  travel  in 
one  day,   (24  hours  ?)  how  many  miles  in  one  year,  (365 
days  ?)  and  how  many  days  would  it  take  a  man  to  travel  this 
last  distance,  at  the  rate  of  40  miles  a  day  ?  how  many  years? 

Ans.  to  the  last,  40800  years. 


1T4R 


MISCELLANEOUS  EXERCISES 


69 


Problems  in  the  Measurement  of  Rectangles 
and  Solids. 


NOTE.  —  A  rectangle  is  a  figure  having  four  sides,  and  each  of  the 
four  corners  a  square  corner. 

PROBLEM  I. 

*ff  48.     The  length  a?id  breadth  of  a  rectangle  given,  to 
find  the  square  contents. 

1.    How  many  square  rods  in  a  plat  of  ground  5  rods  long 
and  3  rods  wide  ? 


SOLUTION.  —  A  square  rod  is  a 
square  measuring  1  rod  on  each  side, 
like  one  of  those  in  the  annexed  dia 
gram.  We  see  from  the  diagram  that 
there  are  as  many  squares  in  a  row  as 
there  are  rods  on  one  side,  and  as  many 
rows  as  there  are  rods  on  the  other 
side  ;  that  is,  5  rows  of  3  squares  in  a 
row,  or  3  rows  of  5  squares  in  a  row. 
We  multiply  the  number  of  squares  in 
one  row  by  the  number  of  rows ;  5  X  3  =  15  square  rods,  Ans. 
Hence  the 

RULE. 

Multiply  the  length  by  the  breadth,  and  the  product  will  be 
the  square  contents. 

NOTE. — Three  times  a  line  5  rods  long  is  a  line  15  rods  long. 
Hence  the  pupil  must  not  fail  to  notice,  that  we  multiply  the  number 
of  square  rods  in  a  piece  of  ground  1  rod  wide  and  of  the  given  length 
by  the  number  of  rods  in  the  width. 

EXAMPLES. 

2.  How  many  square  rods  in  a  piece  of  ground   160  rods 
long  (squares  in  a  row)  and  8  rods  wide  (rows  of  squares)  ? 

Ans.  1280  square  rods. 

3.  How  many  square  feet  in  a  floor  32  feet  long  and  23 
feet  wide  ?  Ans.  736. 

4.  How  many  yards  of  carpeting,  1  yard  wide,  will  it  take 

Questions.  —  ^[  48.  Describe  a  rectangle  ;  a  square  rod.  How 
do  you  determine  the  number  of  squares  in  a  row,  and  the  number  of 
rows  ?  Give  the  r  lie.  What  is  the  quantity  really  multiplied  ?  Wha* 
absurdity  in  considering  it  otherwise  ? 


70  MISCELLANEOUS  EXERCISES.  1149,50. 

to  cover  the  floors  of  two  rooms,  one  8  yards  long  and  7  yards 
wide,  and  the  other  6  yards  long  and  5  yards  wide  ? 

Ans.  86  yards. 

5.  How  many  square  feet  of  boards  will  it  take  for  the 
floor  of  a  room  16  feet  long  and  15  feet  wide,  if  we  allow  12 
square  feet  for  waste  ?  Ans.   252. 

6.  There  is  a  room  6  yards  long  and  5  yards  wide ;  how 
many  yards  of  carpeting,  a  yard  wide,  will  be  sufficient  to 
cover  the  floor,  if  the  hearth  and  fireplace  occupy  3  square 
yards?  Ans.  27. 

PROBLEM  II. 

5T  49.  The  square  contents  and  width  given,  to  find  the 
length. 

1.  What  is  the  length  of  a  piece  of  ground  3  rods  wide, 
and  containing  15  square  rods  ? 

SOLUTION. — In  this  example  we  have  15,  the  number  of  squares 
m  several  rows,  (see  the  diagram,  problem  I.,)  and  3  the  number  of 
squares  in  1  row,  to  find  the  number  of  rows.  We  divide  the  squares 
in  the  number  of  rows  by  the  squares  in  1  row.  Hence, 

RULE. 

Divide  the  square  contents  by  the  width,  and  the  quotient 
will  be  the  length.  Or  really,  since  the  divisor  and  dividend 
must  be  of  the  same  denomination,  we  divide  the  whole  num 
ber  of  square  rods  by  the  square  rods  in  a  piece  of  land  3  rods 
long  by  1  rod  wide ;  thus,  15  -i-  3  =  5  rods  in  length,  Ans. 

EXAMPLES. 

2.  A  piece  of  ground  containing   1280  square  rods,  is  8 
rods  in  width  ;  what  is  its  length  ?  Ans.  160  rods. 

3.  A  floor  containing  736  square  feet,  is  23  feet  wide ; 
what  is  its  length  ?  Ans.  32  feet. 

PROBLEM  III. 

IT  5O.  The  square  contents  and  length  given,  to  find  the 
ividth. 

1.  What  is  the  width  of  a  piece  of  ground,  5  rods  long, 
and  containing  15  square  rods  ? 

Questions.  — 1[49.  Repeat  the  2d  problem ;  the  example.  What 
two  things  are  given  in  the  example,  and  what  required?  Give  th« 
rule.  What  is  really  the  divisor,  and  why  ? 


151.  MISCELLANEOUS  EXERCISES.  71 

SOLUTION. — We  divide  the  square  contents  by  the  length,  or 
really  by  ihe  square  contents  of  a  portion  of  the  ground  5  rods  long 
and  1  rod  wide.  Ans.  3  rods. 

Hence, 

RULE. 

Divide  the  square  contents  by  the  length,  and  the  quotient 
will  be  the  width. 

EXAMPLES. 

2.  A  piece  of  ground  containing  1280  square  rods,  is  160 
rods  in  length  ;  what  is  its  width  ?  Ans.  S  rods. 

3.  What  is  the  width  of  a  field  186  rods  long,  and  con 
taining  13392  square  rods  ?  Ans.  72  rods. 

PROBLEM  IV. 

IT  51.  The  length,  breadth,  and  hight,  or  thickness  given, 
to  find  the  contents  of  a  solid  body.* 

1.  How  many  solid  feet  of  wood  in  a  pile  5  feet  long,  3 
feet  wide,  and  4  feet  high  ? 

SOLUTION.  —  A  solid  foot  is  a 
solid  1  foot  long,  1  wide,  and  1 
high.  By  carefully  inspecting  the 
diagram,  we  may  see  that  a  portion 
of  wood  5  feet  long,  1  foot  wide,  and 
1  high,  will  contain  5  solid  feet. 
Multiplying  5  solid  feet  by  3,  we 
get  the  contents  of  a  portion  5  feet 
long,  3  feet  wide,  and  1  foot  high. 
5  X  3  =  15  solid  feet  ;  and  multi 
plying  15  solid  feet  by  4,  we  get  the  contents  of  the  wrhole  pile,  15  X 
4  =  60  solid  feet,  Ans.  These  are  the  quantities  multiplied,  but  for 
convenience  we  adopt  the  following 

RULE. 

Multiply  the  length  by  the  breadth,  and  the  resulting  prod 
uct  by  the  hight. 

EXAMPLES. 

2.  A  laborer  engaged  to  dig  a  cellar  27  feet  long,  21  feet 

Questions.  —  1  50.  Repeat  the  3d  problem j  the  example;  rule. 
What  is  really  the  divisor,  and  why  ? 

1  51.  What  is  the  4th  problem?  the  first  example?  Describe  a 
solid  foot.  What  quantity  do  you  multiply  in  the  first  multiplication  ? 
in  the  second  ?  What  rule  do  you  adopt  for  convenience  ? 

*  The  cube,  or  right  prism. 


72  MISCELLANEOUS  EXERCISES,  152. 

wide,  and  6  feet  deep;  how  many  solid  feet  must  he  re 
move  ?  A?is.  3402  solid  feet. 

3.  A  farmer  has  a  mow  of  hay  28  feet  long,  14  feet  wide, 
and  8  feet  high ;  how  man%  solid  feet  does  it  contain  ? 

Ans.  3136  solid  feet. 

PROBLEM  V. 

IT  52.  The  solid  contents,  length,  and  breadth  given,  to 
find  the  hight 

1.  A  pile  of  wood  5  feet  long  and  3  feet  wide,  contains  60 
solid  feet ;  what  is  its  hight  ? 

SOLUTION.  —  Since  the  divisor  must  be  of  the  same  denomination 
as  the  dividend,  (solid  feet,)  we  have  given  the  solid  contents  of  a  pile 
5  feet  long,  3  feet  wide,  and  several  feet  high,  which  we  divide  by 
the  solid  contents  of  a  portion  having  the  same  length  and  breadth, 
and  1  foot  high,  to  get  the  number  of  feet  in  the  hight  of  the  pile. 
Thus,  5X3=15,  and  GO  -;-  15  =  4  feet  in  hight,  Ans.  Hence, 

RULE. 

Divide  the  solid  contents  by  the  product  of  the  length  mul 
tiplied  by  the  breadth. 

EXAMPLES. 

2.  A  man  dug  a  cellar  27  feet  long,  and  21  feet  wide,  and 
removed  3402  solid  feet  of  earth ;  what  was  its  depth  ? 

Ans.  6  feet. 

3.  A  mow  of  hay,  28  feet  long  and  14  feet  wide,  contains 
3136  solid  feet;  what  is  its  hight?  Ans.  8  feet. 

NOTE. — Tn  a  similar  manner  we  may  find  the  breadth  or  the 
length,  when  the  solid  contents  and  the  other  two  dimensions  are 
given. 

4.  A  pile  of  wood,  4  feet  wide  and  6  feet  high,  contains 
360  solid  feet ;  what  is  its  length  ?  Ans.   15  feet. 

5.  A  stick  of  timber,  78  inches  long  and  8  inches  thick, 
contains  6864  solid  inches ;  what  is  its  width  ? 

Ans.  11  inches. 

Questions.  —  If  52.  What  is  the  5th  problem  ?  the  first  example  ? 
solution?  rule?  When  the  solid  contents,  width,  and  hight  are  given, 
how  may  the  length  be  found?  When  the  solid  contents,  length,  and 
hight  are  given,  how  may  the  width  be  found  ? 


fl  53-55.  MISCELLANEOUS  EXERCISES.  73 

IT  53.  General  questions  to  be  answered  mentally,  or  by 
the  slate 

If  the  number  of  squares  be  84,  and  the  squares  in  a  row 
be  14,  how  many  will  be  the  rows  of  squares  ? 

If  the  number  of  squares  be  9500,  and  the  rows  of  squares 
be  76,  how  many  will  be  the  squares  in  a  row  ? 

Were  you  required  to  form  an  oblong  field  containing  96 
square  rods,  what,  and  how  many  ways  might  you  vary  the 
figure,  (rows  of  squares  and  squares  in  a  row,)  each  figure  to 
contain  just  96  square  rods  ? 

There  is  a  frame,  40  feet  square  and  18  feet  high,  the  sides 
of  which  are  to  be  covered  with  boards  13  feet  long,  1  foot 
wide  ;  what  number  of  these  boards  will  it  take,  allowing  only 
7  feet  waste  ?  Ans.  222  boards. 

A  room,  in  a  furniture  warehouse,  is  36  feet  long  and  29 
feet  wide ;  how  many  tables,  3  feet  square,  can  be  set  in  it, 
leaving  a  space  2  feet  wide  on  one  of  the  sides  ? 

Ans.  108  tables. 


IT  54.    Definitions. 

Integers  are^  distinguished  as  prime,  composite,  even,  and 
odd. 

1.  A  Prime  number  is  one  that  cannot  be  divided  by  any 
number  except  itself  and  unity  without  a  remainder;  as,  1, 
2,  3,  5,  7,  11. 

NOTE.  —  Two  numbers  are  prime  to  each  other,  as  8  and  15,  when 
a  unit  is  the  only  number  by  which  both  of  them  can  be  divided. 

2.  A  Composite  number ',  see  IT  24. 

3.  An  Even  number  is  one  which  is  exactly  divisible  by  2. 

4.  An  Odd  number  is  one  which  is  not  exactly  divisible 
by  2. 

IT  55.  1.  One  number  is  a  Measure  of  another  whsn  it 
divides  it  without  a  remainder.  Thus,  2  is  a  measure  of  18 ; 
5  of  45;  16  of  64. 

2.  A  number  is  a  Common  Measure  of  two  or  more  num 
bers  when  it  divides  each  of  them  without  a  remainder.  Thus 
3  is  a  common  measure  of  6  and  18;  7  of  28  and  42;  4  of 
12,  20  and  32;  5  of  10,  15,  20,  25. 

Questions.  —  If  54.  How  are  integers  distinguished  ?  What  is  9 
prime  number  ?  composite  number  ?  even  number  ?  odd  number  ? 


74  MISCELLANEOUS  EXERCISES.  IT  56,  57 

3.  One  number  is  a  Multiple  of  another  when  it  can  be 
divided  by  it  without  a  remainder.     Thus  8  is  a  multiple  of 
2;   15  of  5;  33  of  11. 

4.  A  number  is  a  Common  Multiple  of  two  or  more  num 
bers  when  it  can  be  divided  by  each  of  them  without  a  re 
mainder.     Thus,  15  is  a  common  multiple  of  3  and  5 ;   16  of 
2,  4  and  8;  28  of  4  and  7 ;  54  of  2,  3,  6,  9,  18  and  27. 

5.  An  Aliquot,  or  even  part,  is  any  number  which  is  con 
tained  in  another  number  exactly  2,  3,  4,  5,  &c.,  times.    Thus, 
3  is  an  aliquot  part  of  15,  so  also  is  5.     Each  of  the  numbers, 
2,  3,  4,  6,  S,  and  12,  is  an  aliquot  part  of  24. 

6.  The  Reciprocal  of  a  number  is  a  unit,  or  1,  divided  by 
the  number.     Thus,  J  is  the  reciprocal  of  2  ;  J  of  3 ;  J  of  4 ; 
i  of  9,  &c. 


IT  56.    General  Principles  of  Division. 

The  value  of  the  quotient  in  division  evidently  depends 
on  the  relative  values  of  the  dividend  and  divisor. 

EXAMPLE. — Let  the  dividend  be  24,  the  divisor  6,  and  the 
quotient  will  be  4.  Multiplying  the  dividend  by  2,  we  in 
effect  multiply  the  quotient  by  2.  Thus,  24  X  2  =  48,  and 
48  -7-  6  =  8,  which  is  2  times  4,  the  quotient  of  24  H-  6. 

Again,  dividing  the  divisor  by  2,  we  in  effect  multiply  the 
quotient  by  2.  Thus,  6  -r-  2  =  3,  and  24  -f-  3  =  8,  which  is 
2  times  4,  the  quotient  of  24  -r-  6,  the  same  as  before.  Hence, 

PRINCIPLE  I.  Multiplying  the  dividend,  or  dividing  the 
divisor,  by  any  number,  is  in  effect  multiplying  the  quotient 
by  that  number. 

^T  37.  Example  as  above,  namely,  dividend  24,  divisor  6, 
and  quotient  4.  Dividing  the  dividend  by  2,  we  in  effect 
divide  the  quotient  by  2.  Thus,  24-5-2=12,  and  12 -j- 6 
=  2,  which  is  equal  to  1  half  of  the  quotient  of  24  -r-  6. 

Again,  multiplying  the  divisor  by  2,  we  in  effect  divide  the 
quotient  by  2.  Thus,  6  X  2  =  12,  and  24  -r-  12  =  2,  which 
is  equal  to  1  half  of  the  quotient  of  24  -r-  6,  the  same  as  be 
fore.  Hence, 

Questions,  —  IF  55.  What  is  a  measure  ?  common  measure  ?  mul 
tiple?  common  multiple?  an  aliquot  part?  the  reciprocal  of  a  quantity  ? 

1156.  On  what  does  the  value  of  the  quotient  in  division  depend? 
What  is  the  1st  prineipie  ? 


f  58-60.  MISCELLANEOUS  EXERCISES.  75 

PRINCIPLE  II.  Dividing  the  dividend,  or  multiplying  the 
divisor  by  any  number,  is  in  effect  dividing  the  quotient  by 
that  number. 


Example,  the  same  as  before.  Multiplying  both 
dividend  and  divisor  by  2  does  not  alter  the  quotient.  Thus, 
24  X2  =  48;  6  X  2  =  12;  and  48  -f-  12  =  4,  which  is  equal 
to  the  quotient  of  24  •—  6. 

Again,  dividing  both  dividend  and  divisor  by  2  does  not 
alter  the  quotient.  Thus,  24  -j-  2  =  12  ;  6-r-2  =  3;  and 
12-;-  3  =  4,  which  is  equal  to  the  quotient  of  24  -r-  6,  the 
same  as  before.  Hence, 

PRINCIPLE  III.  Multiplying  or  dividing  both  dividend  and 
divisor  by  the  same  number  does  not  alter  the  quotient. 

IT  59.  EXAMPLE.  It  is  required  to  multiply  24  by  6,  and 
divide  the  product  by  6.  24  X  6  =  144,  and  the  product 
]  44  -7-  6  =  24,  which  is  equal  to  the  number  multiplied. 
Hence, 

PRINCIPLE  IV.  If  a  number  be  multiplied,  and  the  product 
divided  by  the  same  number,  the  quotient  will  be  the  number. 

This  result  depends  upon  the  principle  that  if  the  product 
be  divided  by  the  multiplier,  the  quotient  will  be  the  multi 
plicand. 


1T6O.    Cancelation. 

1.  How  many  oranges,  at  4  cents  apiece,  can  be  bought 
for  4  dimes,  or  4  ten  cent  pieces  ? 

SOLUTION.  —  We  multiply  10  by  4  to  get  the  number  of  cents, 
10  X  4  =  40  ;  then  as  many  times  as  4  is  contained  in  40  so  many 
oranges  can  be  bought.  But  multiplying  10  and  dividing  the  pro 
duct  by  the  same  number  does  not  change  it,  (^[59  ;)  hence,  we  maj 
omit  both  operations,  taking  10  for  the  result,  as  follows  : 

OPERATION.  Writing  10  and  the  multiplier  4  above,  and  the 

10  V  4  divisor  4  below  a  horizontal  line,  we  strike  out  4 

^      _ — .  1Q     above  and  below  the  line,  and  we  have  10  for  the 
^  result.  Ans.  10  oranges. 

NOTE.  —  This  process  of  omitting  4  is  called  cancelation.  When 
we  cancel  a  number,  we  usually  draw  an  oblique  line  across  it. 

Questions.  —  f  57.    What  is  the  2d  principle  ? 
If  58.    What  is  the  3d  principle  ? 
if  59.     What  is  the  4th  principle  ? 


76  MISCELLANEOUS  EXERCISES  II  60. 

2.  A  farmer  sold  15  cows  for  24  dollars  apiece,  and  took 
his  pay  in  sheep  at  5  dollars  apiece  ;  how  many  sheep  did 
he  receive  ? 

SOLUTION.  —  We  see  that  24  is  to  be  multiplied  by  the  composite 
number  15  =  3  X  5,  and  the  product  divided  by  5.  Using  the  com 
ponent  parts  of  the  multiplier,  we  multiply  24  by  3.  Now  the  product 
of  24  X  3  is  to  be  multiplied  and  the  result  divided  by  5,  which  oper 
ations  we  may  omit,  as  follows  : 

Writing  the  numbers  as  already  described,  we 

OPERATION.        strike  out  5  below,  and  15  =  3  X  5  above  the  line, 

and  above  15  set  the  factor  3,  by  which  we  multi- 

24  X  X&          _     ply  24.     Since  there  is  no  number  by  which  to  di- 

-T        =  12    vjje  ^1,3  prOQUCt  it  is  the  result  required. 

Ans.  72  sheep. 

3.  Multiply  165  by  33,  and  divide  the  product  by  31; 
multiply  the  quotient  by  16  and  divide  the  product  by  99  ; 
multiply  the  quotient  by  62  and  divide  the  product  by  55  ; 
multiply  the  quotient  by  3  and  divide  the  product  by  20. 

OPERATION.  By    closely    in- 

g  42  specting  these  num- 


#2>C09XjM»X!ff         5  5    the  line  are  canceled 

0  5  except  4,  2  and  3, 

which  must  be  mul 

tiplied  together  ;  and  that  all  the  factors  below  the  line  are  canceled 
except  5,  by  which  the  product  of  the  remaining  factors  above  the  line 
is  to  be  divided. 

NOTE  1.  —  It  is  plain  that  16  above  and  20  below  the  line  have  the 
factor  4  common,  for  16  =  4  X  4  and  20  =  4  X  5  ;  we  therefore  can 
cel  the  factor  4  from  16  and  20  ;  this  we  do  if  we  erase  the  two  num 
bers,  and  write  4  the  other  factor  of  16  over  it,  and  5  the  other  factor 
of  20  under  it.  We  see  also  that  3,  the  reserved  factor  of  165,  can 
eels  3,  the  reserved  factor  of  99. 

NOTE  2.  —  If  the  pupil  will  perform  the  operations  at  length,  of 
multiplying  and  dividing,  in  this  example,  he  will  see  how  much  is 
saved  by  cancelation. 

Cancelation,  then,  is  the  method  of  erasing,  or  rejecting,  a 
factor  or  factors,  from  any  number  or  numbers.  It  may  be 
applied  for  shortening  the  operation  where  both  multiplication 
and  division  are  required,  by  rejecting  equal  factors  from  the 
numbers  to  be  multiplied  and  the  divisors. 


IT  60.  MISCELLANEOUS  EXERCISES.  77 

RULE. 

I.  Write  down  the  numbers  to  be  multiplied  together  above, 
and  the  divisors  below,  a  horizontal  line. 

II.  Cancel  all  the  factors  common  to  the  numbers  to  be 
multiplied  and  the  divisors. 

III.  Proceed  with  the  remaining  numbers  as  required  by 
the  question. 

NOTE.  —  One  factor  on  one  side  of  the  line  will  cancel  only  one  like 
factor  on  the  other  side. 

EXAMPLES  FOR  PRACTICE. 

4.  A  man  sold  35  barrels  of  flour  at  5  dollars  per  barrel,  and 
took  his  pay  in  salt  at  3  dollars  per  barrel ;  he  sold  the  salt  at 
4  dollars  per  barrel,  and  took  his  pay  in  broadcloth  at  7  dol 
lars  per  yard ;  he  sold  the  broadcloth  at  8  dollars  per  yard, 
and  took  his  pay  in  sheep  at  2  dollars  a  head ;  he  sold  the 
sheep  at  3  dollars  a  head,  and  took  his  pay  in  land  at  15  dol 
lars  per  acre  ;  how  many  acres  of  land  did  he  purchase  ? 

If  like  factors  be  canceled  from  the  numbers  to  be  multi 
plied  and  the  divisors,  there  will  remain  of  the  numbers  to  be 
multiplied  5  X  4  X  4  =  80,  and  of  the  divisors  3 ;  and  *-£-  = 
26|.  Am.  26 §  acres. 

5.  What  is  the  quotient  of36xSx4x8x2  divided  by 
6X5X3X4X2? 

NOTE.  —  The  remaining  factors  of  the  numbers  to  be  multiplied 
are  2,  8  and  8,  and  of  the  divisors,  5. 

6.  In  a  certain  operation  the  numbers  to  be  multiplied  are 
27,  14,  40,  8  and  6,  and  the  divisors  are  7,  10,  12  and  15 ; 
what  is  the  quotient  ? 

9X2X2  X  8=288,  and  288 -4- 5  =  57-f ,  Am. 

7.  What  is  the  quotient  of  4  X  7  X  18  X  10  X  8  X  9, 
divided  by  24  X  72  X  3  ? 

NOTE.  —  All  the  divisors  cancel.     Ans.  70. 

8.  If  the  numbers  to  be  multiplied  are  14,  5,  3  and  28,  and 
the  divisors  15  and  9 ;  what  is  the  quotient  ? 

NOTE.  — The  remaining  factor  of  the  divisors  is  9.      Ans.  43 ^. 


Questions.  —  1[6O.     If  a  number  be  multiplied  and  the  product 

divided  by  the  same  number,  what  is  the  result?  When  such  operations 
are  to  be  performed,  how  may  they  be  contracted?  What  is  this  pro 
cess  called?  How  do  you  indicate  that  a  number  is  canceled?  What 
is  cancelation ?  When  may  it  be  applied?  Repeat  tne  rule.  Explain 
the  operation  in  Ex.  5  ;  in  Ex.  6,  &e. 
7* 


78  MISCELLANEOUS  EXERCISES.  IT  61,  62. 

5T61.     To  find  a  co?nmon  divisor  of  two  or  more  numbers. 

1.  Find  a  common  divisor  of  6,  9  and  12. 

!6  =  3  X  2        The  factor  3,  which  is  common  to  the 
9  =  3X3     several  numbers,  must  be  a  common  divi- 
12  =  3  X  4     sor  of  them.     Hence  the 

RUL.E. 

Separate  each  number  into  two  factors,  one  of  which  shall 
be  common  to  all  the  numbers. 

The  common  factor  will  be  their  common  divisor. 

EXAMPLES  FOR  PRACTICE. 

2.  Find  a  common  divisor  of  4,  16,  24,  36  and  8. 

Ans.  4. 

3.  Find  a  common  divisor,  or  common  measure,  (which 
terms  mean  the  same  thing,)  of  22,  44,  66,  and  88. 

Ans.  11. 

4.  Required  the  length  of  a  rod  which  will  be  a  common 
measure  of  two  pieces  of  cloth,  one  of  them  25  feet,  the  other 
30  feet  long.  Ans.  5  feet. 


To  find  the  greatest  common  divisor  of  two  or  more 
numbers. 

The  greatest  common  divisor  of  several  numbers  is  the 
greatest  factor  common  to  them,  and  may  be  found  by  a  sort 
of  trial.  Let  it  be  required  to  find  the  greatest  common  divi 
sor  of  128  and  160.  The  greatest  common  divisor  cannot 
exceed  the  less  number,  for  it  must  measure  it.  We  will  try, 
therefore,  if  the  less  number,  128,  which  measures  itself,  will 
also  divide  or  measure  160. 

lOQMfin  /  1  1^8  in  160,  1  time,  and  32  remain;  128, 

128  therefore,  is  not  a  divisor  of  160.     We  will 

now  try  whether  this  remainder  be  not  the 

32  }  128  (4     Divisor  sought;  for  if  32  be  a  divisor  of 

'  1  go  ^        128,  the  former  divisor,  it  must  also  be  a 

divisor  of  160,  which  consists  of  128  -fc-  32. 

32  in  128,  4  times,  without  any  remainder. 

Consequently  it  is  contained  in  160  =  128  -\-  32,  just  5  times; 

that  is,  once  more  than  in  128.     And  as  no  number  greater 

than  32,  the  difference  of  the  two  numbers,  is  contained  once 

more  in  the  greater,  it  is  the  greatest  common  divisor.    Hence, 

Questions.  —  IT  61.  What  is  a  common  divisor  of  two  or  more 
numbers  ?  Repeat  the  rule  for  finding  it. 


1T62.  MISCELLANEOUS  EXERCISES.  79 

r~ 

To  find  the  greatest  common  measure  of  two  numbers, 

RULE. 

Divide  the  greater  number  by  the  less,  and  that  divisor  by 
the  remainder,  and  so  on,  always  dividing  the  last  divisor  by 
the  last  remainder,  till  nothing  remain.  The  last  divisor  will 
be  the  greatest  common  divisor  required. 

NOTE  1.  —  When  we  would  find  the  greatest  common  divisor  ol 
more  than  two  numbers,  we  may  first  find  the  greatest  common  divi 
sor  of  two  numbers,  and  then  of  that  common  divisor  and  one  of  the 
other  numbers,  and  so  on  to  the  last  number.  Then  will  the  greatest 
common  divisor  last  found  be  the  answer. 

NOTE  2.  —  Two  numbers  which  are  prime  to  each  other,  of  course, 
can  have  no  common  divisor  greater  than  r. 

EXAMPLES    FOR    PRACTICE. 

1.  Apply  the  foregoing  rule  to  find  the  greatest  common 
divisor  of  21  and  35. 

2.  Find  the  greatest  common  divisor  of  96  and  544. 

Ans.  32. 

3.  Find  the  greatest  common  divisor  of  468  and  1 184. 

Ans.  4. 

4.  What  is  the  greatest  common  divisor  of  32,  SO,  and 
256?  Ans.  16. 

5.  What  is  the  greatest  common  divisor  of  75,  200,  625, 
and  150  ?  Ans.  25. 

6.  A  certain  tract  of  land  containing  100  acres,  is  160  rods 
long  and  100  wide ;  what  is  the  length  of  the  longest  chain 
that  will  exactly  measure  both  its  length  and  breadth  ? 

Ans.  20  rods. 

7.  A  has  2640  dollars,  B  1680  dollars,  and  C  756  dollars, 
which  they  agree  to  lay  out  for  land  at  the  greatest  price  per 
acre  that  will  allow  each  to  expend  the  whole  of  his  money ; 
what  was  the  p'rice  per  acre,  and  how  many  acres  did  each 
man  buy  ? 

Ans.  A  bought  220  acree,  B  140  acres,  and  C  63  acres,  at 
12  dollars  per  acre. 

Questions.  —  If  62.  What  is  the  greatest  common  divisor  of  two 
or  more  numbers  ?  Describe  the  process  of  finding  it  for  two  numbers  ? 
rule?  How  found  when  the  numbers  are  more  than  two?  What  is  tb- 
greatest  common  measure  of  numbers  that  are  prime  to  each  ether  ? 


80  COMMON  FRACTIONS.  163. 

« 

COMMON    FRACTIONS. 

IT  63.  When  whole  numbers,  which  are  called  integers, 
(IT  36,)  are  subjects  of  calculations  in  arithmetic,  the  opera 
tions-are  called  operations  in  whole  numbers.  But  it  is  often 
necessary  to  make  calculations  in  regard  to  parts  of  a  thing  or 
unit.  We  may  not  only  have  occasion  to  calculate  the  price 
of  3  barrels,  5  barrels,  or  8  barrels  of  flour,  but  of  one  third 
of  a  barrel,  two  fifths  of  a  barrel,  or  seven  eighths  of  a  barrel. 

When  a  unit  or  whole  thing  is  divided  or  broken  into  any 
number  of  equal  parts,  the  parts  are  called  fractions,  OT  broken 
numbers,  (from  the  Latin  word,  frango,  I  break.)  If  it  be 
divided  into  3  equal  parts,  the  parts  are  called  thirds  ;  if  into 
7  equal  parts,  sevenths ;  if  into  12  equal  parts,  twelfths.  The 
fraction  takes  its  name,  or  denomination,  from  the  number  of 
parts  into  which  the  unit  or  whole  thing  is  divided. 

If  the  unit  or  whole  thing  be  divided  into  16  equal  parts, 
the  parts  are  called  sixteenths,  and  5  of  these  parts  would  be 
5  sixteenths. 

Fractions  are  of  three  kinds,  Common,  (sometimes  called 
Vulgar,)  Decimal,  and  Duodecimal. 

Common  fractions  are  always  expressed  by  two  numbers, 
one  above  the  other,  with  a  horizontal  line  between  them ; 
thus,  J,  |,  f 

The  number  below  the  line  is  called  the  Denominator,  be 
cause  it  gives  name  to  the  parts. 

The  number  above  the  line  is  called  the  Numerator,  because 
it  numbers  the  parts. 

The  denominator  shows  into  how  many  parts  a  thing  or 
unit  is  divided  ;  and 

The  numerator  shows  how  many  of  these  parts  are  con- 
tamed  in  the  fraction.  Thus,  in  the  fraction  f ,  the  denomina 
tor,  8,  shows  that  the  unit  or  whole  thing  is  divided  into  8 
equal  parts,  and  the  numerator,  3,  shows  that  3  of  these  parts 
are  contained  in  the  fraction.  The  numerator,  3,  numbers  the 
parts ;  the  denominator,  8,  gives  them  their  denomination  or 

Questions.  —  If  63.  What  are  integers?  What  fractions,  and 
whence  their  necessity  ?  Whence  do  fractions  take  their  name  ?  How 
many  kinds  of  fractions  ?  Name  them.  How  are  common  fractions 
written  ?  What  is  the  lower  number  called,  and  why  ?  What  does  it 
show  ?  What  is  the  upper  number  called,  and  why  ?  What  determines 
the  size  of  the  parts,  and  why  ?  What  are  the  terms  of  a  fraction  ? 
What  are  the  terms  of  the  fraction  -fa  ?  -££  ?  &c. 


1T64,  66.  COMMON  FRACTIONS.  81 

name,  and  shows  their  size  or  magnitude ;  for  if  a  thing  be 
divided  into  8  equal  parts,  the  parts  are  but  half  as  large  as 
if  divided  into  but  4  equal  parts.  It  will  evidently  take  2 
eighths  to  make  1  fourth. 

The  numerator  and  denominator,  taken  together,  are  called 
the  terms  of  the  fraction.  Thus,  the  terms  of  the  fraction.  ^ 
are  7  and  10 ;  of  §,  2  and  8. 

IT  64.  It  is  important  to  bear  in  mind,  that  fractions  arise 
from  division,  and  that  the  numerator  may  be  considered  a 
dividend,  and  the  denominator  a  divisor,  and  the  value  of  the 
fraction  the  quotient ;  thus,  \  is  the  quotient  of  1  (the  nu 
merator)  divided  by  2,  (the  denominator ;)  £  is  the  quotient 
arising  from  1  divided  by  4 ;  and  f  is  3  times  as  much,  that 
is,  3  divided  by  4;  thus,  1  fourth  part  of  3  is  the  same  as  3 
fourths  of  1. 

Hence,  a  common  fraction  is  always  expressed  by  the  sign 
of  division,  the  numerator  being  written  in  the  place  of  the 
upper  dot,  and  the  denominator  in  the  place  of  the  lower  dot. 


|  expresses  the  quotient,  of  which  \  f 


dend  or  numerator, 
.sor  or  denominator. 


1.  If  4  oranges  be  equally  divided  among  6  boys,  what 
part  of  an  orange  is  each  boy's  share  ? 

A  sixth  part  of  1  orange  is  £,  and  a  sixth  part  of  4  oranges 
is  4  such  pieces,  =  f.  Ans.  f  of  an  orange. 

2.  If  3  apples  be  equally  divided  among  5  boys,  what  part 
of  an  apple  is  each  boy's  share  ?  if  4  apples,  what  ?  if  2  ap 
ples,  what?  if  5  apples,  what? 

3.  What  is  the  quotient  of  1  divided  by  3  ?  of  2  by 

3  ? of  1  by  4  ? of  2  by  4  ? of  3  by  4  ? of 

5  by  7?  of  6  by  8?  of  4  by  5? of  2  by  14? 

4.  What  part  of  an  orange  is  a  third  part  of  2  oranges  ? 

one  fourth  of  2  oranges  ?  J  of  3  oranges  ?  -£ 

of  3  oranges?  iof  4?  *of  2?  4  of  5?  

f  Of3?  £of  2? 

IT  65.  A  fraction  being  part  of  a  whole  thing,  is  properly 
less  than  a  unit,  and  the  numerator  will  be  less  than  the  de 
nominator,  since  the  denominator  shows  how  many  parts 

Questions.  —  If  64.  From  what  do  fractions  always  arise  ?  What 
may  the  numerator  be  considered?  the  denominator?  What  is  the  value 
of  the  fraction  ?  Of  what  is  £  the  quotient  ?  |  ?  -f  £  ?  £  of  3  is  what  part 
of  1  ?  i\j-  of  7  is  what  part  of  1  ?  By  what  is  a  common  fraction  al 
ways  expressed  ? 


82  COMMON  FRACTIONS.  H  65. 

make  a  whole  thing,  and  there  must  not  be  so  many  of  the 
parts  taken  as  will  make  a  whole  thing1. 

But  we  call  an  expression  written  in  the  fractional  form 
a  fraction,  though  its  numerator  equals  or  exceeds  the  denom 
inator,  and  its  value,  consequently,  equals  or  exceeds  a  unit ; 
but  since  there  is  not  a  strict  propriety  in  the  name,  it  is  called 
an  improper  fraction.  Hence, 

A  Proper  Fraction  is  one  that  is  less  than  a  unit,  its  nu 
merator  being  less  than  the  denominator. 

An  Improper  Fraction  is  one  that  equals  or  exceeds  a  unit, 
its  numerator  equaling,  or  exceeding  the  denominator.  Thus, 
JjA,  -||-,  are  improper  fractions. 

A  Simple  Fraction  is  a  single  fraction,  either  proper  or  im 
proper.  Thus,  |,  f ,  -ff ,  are  simple  fractions. 

A  Compound  Fraction  is  a  fraction  of  a  fraction,  or  several 
fractions  connected  by  the  word  of.  Thus,  J  of  f ,  -f  of -y^  f 
of  J  of  -^-,  are  compound  fractions. 

A  Complex  Fraction  is  one  which  has  a  fraction,  either 
simple  or  compound,  or  a  mixed  number,  for  its  numerator, 

2    43    2.  Of  4 
or  for  its  denominator,  or  for  both.     Thus,  ^-,  — -,  s         ,  are 

^f    F       ^f 
complex  fractions. 

A  Mixed  Number,  as  already  shown,  is  one  composed  of  a 
whole  number  and  a  fraction.  Thus,  14J,  13£,  &c.,  are 
mixed  numbers. 

A  father  bought  4  oranges,  and  cut  each  orange  into  6 
equal  parts ;  he  gave  to  Samuel  3  pieces,  to  James  5  pieces, 
to  Mary  7  pieces,  and  to  Nancy  9  pieces ;  what  was  each 
one's  fraction  ? 

Was  James'  fraction  proper  or  improper  ?     Why  ? 

Was  Nancy's  fraction  proper  or  improper  ?     Why  ? 

If  an  orange  be  cut  into  5  equal  parts,  by  what  fraction  is 

1  part  expressed  ?  2  parts  ?  3  parts  ?  4  parts  ? 

' 5  parts  ?  How  many  parts  will  make  unity  or  a  whole 

orange  ? 

If  a  pie  be  cut  into  8  equal  pieces,  and  two  of  these  pieces 
be  given  to  Harry,  what  will  be  his  fraction  of  the  pie  ?  if  5 

Questions.  —  IT  65.  What  is  a  proper  fraction,  and  why  so  called? 
its  value  ?  What  is  an  improper  fraction,  and  why  so  called  ?  When 
is  its  value  a  unit  ?  When  greater  than  a  unit  ?  Why  ?  What  is  a 
simple  fraction  ?  a  simple  proper  fraction  ?  a  simple  improper  fraction  I 
a  compound  fraction?  a  complex  fraction?  a  mixed  number?  What 
kind  of  a  fraction  is  f  of  &  of  -^  ?  More  questions  of  this  character. 


1T66. 


COMMON  FRACTIONS. 


83 


pieces  be  given  to  John,  what  will  be  his  fraction  ?  what  frac 
tion  or  part  of  the  pie  will  be  left  ? 


IT  66.    Reduction  of  Fractions. 

Reduction  of  fractions  is  changing  them  from  one  form  to 
another  without  altering  their  value. 


To  reduce  an  improper  frac 
tion  to  a  whole  or  mixed  num 
ber. 

1.  In  4  halves  (f-)  of  an 
apple  how  many  whole  ap 
ples? 

SOLUTION.  —  Since  2  halves 
(f  )  of  an  apple  are  equal  to  1 
whole  apple,  4  halves  (f-)  are 
equal  to  as  many  apples  as  the 
number  of  times  2  halves  are  con 
tained  in  4  halves,  which  is  2 
times.  Arts.  2  apples. 

3.  In  f  of  an  apple  how 
many  whole  apples  ?  -  in 
f  ? 


n 


5.  How  many  yards  in  § 
of  a  yard?  -  in  f  of  a 
yard?  -  in  f  ?  -  in  §? 


in 
in 
in 


n 


n 


7.  How  many  bushels  in  8 
pecks  ?  that  is,  in  f  of  a  bush 
el  ?  in  Jj0-  ?  ir 

in   J3?    jn 


n 


9.  If  I  give  27  children  J 
of  an  orange  each,  how  many 
oranges  will  it  take  ? 


To  reduce  a  whole  or  mixed 
number  to  an  improper  frac 
tion. 

2.  In  2  whole  apples  how 
many  halves  ? 

SOLUTION. — In  2  apples  are 
two  times  as  many  halves  as  there 
are  in  1  apple.  Since  there  are 
2  halves  (f )  in  1  apple,  there  are 
2  times  2  halves  in  2  apples,  =4 
halves,  that  is,  f,  Ans. 

4.  In  3  apples  how  many 
halves  ?  in  4  apples  ?  in  6  ap 
ples  ?  in  10  apples  ?  in  24 ? 
in  60?  in  170?  in  492? 

6.  Reduce  2  yards  to 
thirds.  Ans.  f .  Reduce  2| 
yards  to  thirds.  Ans.  f.  Re 
duce  3  yards  to  thirds.  

3|  yards.  3|  yards. 

5  yards.  5|  yards. 

6 1  yards. 

8.  Reduce  2  bushels  to 

fourths.  2f  bushels. 

6  bushels. 6J  bush 
els.  7|  bushels.  

25f  bushels. 

10.  In  6j  oranges  how 
many  fourths  of  an  orange  ? 


64 


COMMON  FRACTIONS. 


H66. 


OPERATION.  It  will  take 

4)27  ¥;  and  it  is 

evident,  that 

Ans.  6J  oranges,  dividing  the 

numerator, 

27,  (  =  the  number  of  parts  con 
tained  in  the  fraction,)  by  the  de 
nominator,  4,  (  =  the  number  of 
parts  in  1  orange,)  will  give  the 
number  of  whole  oranges,  and  the 
remainder,  written  over  the  de 
nominator,  will  express  the  frac 
tional  part.  Hence, 

To  reduce  an  improper 
fraction  to  a  whole  or  mixed 
number, 

RULE. 

Divide  the  numerator  by 
the  denominator  ;  the  quo 
tient  will  be  the  whole  or 
mixed  number. 


OPERATION. 
6  j  oranges. 
4 

24  fourths  in  6  oranges. 
3     "   cont'd  in  the  fraction. 

27  = 


Since  there  are  4  fourths  in  I 
orange,  in  6  oranges  there  are 
6  times  4  fourths  =  24  fourths, 
and  24  fourths  -j-  3  fourths  =  27 
fourths.  Hence, 

To  reduce  a  mixed  number 
to  an  improper  fraction, 

RULE. 

Multiply  the  whole  number 
by  the  denominator  of  the 
fraction ;  to  the  product  add 
the  numerator,  and  write  the 
result  over  the  denominator. 

NOTE  1.  — A  whole  number  may  be  reduced  to  the  form  of  an  im 
proper  fraction,  by  writing  1  under  it  for  a  denominator. 

NOTE  2.  —  A  whole  number  may  be  reduced  to  a  fraction  having  a 
specified  denominator,  by  multiplying  the  whole  number  by  the  given 
denominator,  and  taking' the  product  for  a  nu  aerator. 

EXAMPLES    FOR    PRA*  TICE. 

11.  In  -8F3-  of  a  dollar,  how  12.  In  13£  dollars,  how 
many  dollars  ?  many  sixths  of  a  dollar  ? 

13.  In  -^{p-  of  an  hour,  14.  What  is  the  impropei 
how  many  hours  ?  fraction  equivalent  to  23f£ 

hours  ? 

16.  Reduce  730T32-  shillings 
to  an  improper  fraction. 


15.    In  *£fp  of  a  shilling, 
how  many  shillings  ? 


Questions.  —  T[  66.  What  is  reduction  of  fractions?  To  what  is 
the  value  of  a  fraction  equal  ?  What  is  the  rule  for  reducing  an  im 
proper  fraction  to  a  whole  or  mixed  number  ?  a  mixed  number  to  an 
improper  fraction  ?  How  may  a  whole  number  be  reduced  to  the  form 
of  an  improper  fraction  ?  How  to  a  fraction  having  a  specified  denom 
inator  ? 


If  67.  COMMON  FRACTIONS.  6b 


17.  In  -3-£Ji  of  a  day,  how  18.  In  156££  days,  .iow 
many  days  ?  many  24ths  of  a  day  ? 

Am.  i^f  i  =  3761  hours. 

19.  In  J^Z_L  of  a  gallon,  20.  In  342f  gallons,  how 
how  many  gallons  ?  many  4ths  of  a  gallon  ? 

.47^.  JL^LL  of  a  gallon  = 

1371  quarts. 

21.  Reduce  *J,  ^/,  flft,  22.  Reduce  l£ft,  17f£, 
ttrr*.  ¥inf  »  to  whole  or  mixed  8^,  4JV)<&,  and  7f  ^  to  im- 
numbers.  proper  fractions. 

IT  G71.  To  reduce  a  fraction  to  its  loivest  or  most  simple 
terms. 

If  |  of  an  apple  be  divided  into  2  equal  parts,  it  becomes  f  . 
The  effect  on  the  fraction  is  evidently  the  same  as  if  we  had 
multiplied  both  of  its  terms  by  2.  In  either  case,  the  parts 
are  made  2  times  as  MANY  as  they  were  before  ;  but  they  are 
only  HALF  AS  LARGE  ;  for  it  will  take  2  times  as  many  fourths 
to  make  a  whole  one  as  it  will  take  halves  ;  and  hence  it  is 
that  f  is  the  same  in  value  or  quantity  as  \. 

f  is  2  parts  ;  and  if  each  of  these  parts  be  again  divided 
into  2  equal  parts,  that  is,  if  both  terms  of  the  fraction  be  mul 
tiplied  by  2,  it  becomes  f  . 

Now  if  we  reverse  the  above  operation,  and  divide  both 
terms  of  the  fraction  f  by  2,  we  obtain  its  equal,  f  ;  dividing 
again  by  2,  we  obtain  J,  which  is  the  most  simple  form  of  the 
fraction,  because  the  terms  are  the  least  possible  by  which  the 
fraction  can  be  expressed.  Hence.  J==f=:|,  and  the  re 
verse  of  this  is  evidently  true,  that  J  =  |  =  |. 

It  follows,  therefore,  by  multiplying  or  dividing  both  terms 
of  the  fraction  by  the  same  number,  ice  change  its  terms  with 
out  altering  its  value.  (IT  58.) 

The  process  of  changing  f  into  its  equal  J,  is  called  reduc 
ing  the  fraction  to  its  lowest  terms. 

A  fraction  is  said  to  be  in  its  lowest  terms  when  no  number 
greater  than  1  will  divide  its  numerator  and  denominator 
without  a  remainder. 

1.    Reduce         to  its  lowest  terms. 


OPERATION.  ^e  ^n(^'  ky  *"a^»  l^at  4  w^*  exactly 

QN  measure  both  128  and  160,  and,  dividing, 

128        32        4  we  °lian£e  tne  fracti°n  to  its  equal  %%. 

4}  -  —  _  —  .  _  Ans.  Again,  we  find  that  8  is  a  divisor  common 

^160       40        5  to  both  terms,  and,  dividing,  we  reduce 
8 


86  COMMON  FRACTIONS.  IF  67. 

the  fraction  to  its  equal  f-,  which  is  now  in  its  lowest  terms,  for  no 
greater  number  than  1  will  again  measure  them. 

NOTE  1.  —  Any  fraction  may  evidently  be  reduced  to  its  lowest 
terms  by  a  single  division,  if  we  use  the  greatest  common  divisor  of 
the  two  terms.  Thus,  we  may  divide  by  32,  which  we  found  (*|[62) 
to  be  the  greatest  common  divisor  of  128  and  160.  32)  -^f£=f  Ans. 


Hence,  To  reduce  a  fraction  to  its  lowest  terms, 

RULE. 

Divido  both  terms  of  the  fraction  by  any  number  which  will 
divide  them  both  without  a  remainder,  and  the  quotients  thence 
arising  in  the  same  manner,  and  so  on,  till  it  appears  that  no 
number  greater  than  1  will  again  divide  them. 

NOTE  2.  —  A  number  ending  with  a  cipher  is  divisible  by  10.  If 
the.  two  right  hand  figures  are  divisible  by  4,  the  whole  number  is  also. 
A  number  is  divisible  by  2  when  it  ends  with  an  even  number,  and  by 
5  when  it  ends  with  5,  or  0. 

EXAMPLES  FOR  PRACTICE. 

2.  Reduce  ££f  to  its  lowest  terms.  Ans.  £. 

3.  Reduce  £g#,  /^,  4-f-|5  and  f£  to  their  lowest  terms. 

Ans.  fc,  A,  *,  f 

NOTE  3.  —  Let  the  following  examples  be  wrought  by  both  methods  ; 
by  several  divisors,  and  also  by  finding  the  greatest  common  divisor. 

4.  Reduce  £-§•§,  ^9?9T,  |£#,  and  £f£f  to  their  lowest  terms. 

"  Ans.  J,  J,  £,  and  f  . 

5.  Reduce  •£££%  to  its  lowest  terms. 

6.  Reduce  ££$  to  its  lowest  terms. 

7.  Reduce  T4T6g\-  to  its  lowest  terms. 

8.  Reduce  £ff  f  to  its  lowest  terms. 

NOTE.  —  The  reducing  of  a  compound  fraction  to  a  simple  one  will 
be  considered  in  the  multiplication  of  fractions,  where  it  properly  be 
longs.  The  reducing  of  fractions  to  a  common  denominator  will  be 
presented  in  connection  with  addition  and  subtraction  of  fractions,  in 
which  operations  only  it  is  necessary.  The  reducing  of  complex  to 
simple  fractions  will  be  considered  after  the  pupil  shall  be  made 
acquainted  with  the  division  of  fractions,  a  knowledge  of  which  is 
indispensable  to  understanding  the  operation. 

Questions  .  —  If  67.  Give  the  illustration  with  the  half  apple.  Re  - 
verse  the  operation.  What  follows?  What  is  the  process  mentioned, 
and  what  is  it  called  ?  When  is  a  fraction  in  its  lowest  terms  ?  Explain 
Ex.  i.  How  can  a  fraction  be  reduced  by  a  single  division?  Rule. 
Give  the  note  by  which  you  determine  by  what  number  you  divide 


IF  68, 69.  COMMON  FRACTIONS.  87 

Addition  and  Subtraction  of  Fractions. 

COMMON  DENOMINATOR. 

IT  68.  1.  A  boy  gave  to  one  of  his  companions  f  of  an 
orange,  to  another  f ,  to  another  J ;  what  part  of  an  orange 
did  he  give  to  all  ?  |  -(-  |  -j-  J  =  how  much  ? 

SOLUTION.  —  The  adding  together  of  f ,  f  and  J  of  an  orange  is 
the  same  as  the  adding  of  2  oranges,  4  oranges,  and  1  orange,  which 
would  make  7  oranges.  The  8  ite  called  the  common  denominator,  as 
it  is  common  to  the  several  fractions  ;  and  we  write  over  it  the  sum 
of  the  numerators,  to  express  the  answer.  Ans.  £. 

2.  A  boy  had  fa  of  a  dollar,  of  which  he  expended  f^ ; 
what  had  he  left  ? 

SOLUTION.  —  y\j-  of  a  dollar  is  one  dime,  or  ten  cent  piece.  The 
operation,  then,  is  to  subtract  3  ten  cent  pieces  from  7  ten  cent  pieces, 
which  will  leave  4  ten  cent  pieces,  or,  Ans,  y^j. 

3.  J  -|-  |  -)-  1  =  how  much  ?     f  —  J  =  how  much  ? 

4-  ^  +  ifr  +  *  +  »  +  *==k°wmuch?  if  — T3F  = 
how  much  ? 

5.  A  boy,  having  f  of  an  apple,  gave  J  of  it  to  his  sister; 
what  part  of  the  apple  had  he  left  ?  f  —  J  =  how  much  ? 

5T69.  1.  A  boy,  having  an  orange,  gave  f  of  it  to  his 
sister,  and  ^  of  it  to  his  brother ;  what  part  of  the  orange  did 
he  give  away  ? 

SOLUTION.  —  The  fractions  $  and  £  of  an  orange  can  no  more  be 
added  than  3  oranges  and  1  apple,  which  would  make  neither  4  oranges 
nor  4  apples,  as  they  are  of  different  kinds,  (*|f  12.)  But  if  1  orange 
made  2  apples,  the  3  oranges  would  make  6  apples,  and  the  1  apple 
being  added  we  should  have  7  apples.  Now  \  does  make  just  f , 
and  consequently  j  make  f ,  to  which  if  £  be  added  we  shall  have 
the  Ans.  |. 

The  denominator,  4,  of  the  fraction  f ,  is  a  factor  of  8, 
the  denominator  of  the  fraction  £.  And  if  each  term  of  the 
fraction  J  be  multiplied  by  2,  the  remaining  factor  of  8,  it  will 
be  reduced  to  8ths'  (f ,)  without  altering  its  value.  (IF  67.) 
Hence,  if  the  denominator  of  one  fraction  be  a  factor  of  the 
denominator  of  anc  ther  fraction,  and  both  its  terms  be  multi- 

Qnestions.  —  If  68.     Like  what,  is  the  process  of  adding  eighths 
What  is  the  8  called,  and  why  ?     What  is  the  tenth  of  a  dollar? 


88  COMMON  FRACTIONS.  1F70. 

plied  by  the  remaining  factor,  it  will  be  reduced  to  the  same, 
denominator  ivith  the  latter  fraction,  without  altering  its  value. 
(IF  58.)  For  example: 

2.  How  many  12ths  in  f  ? 

SOLUTION.  —  The  factors  of  12  are  3  and  4,  the  latter  of  which  is 
the  denominator  of  f  ,  and  multiplying  both  terms  of  f  by  3,  the  other 
factor,  we  have  -f^-,  a  fraction  of  the  same  value  as  f  ,  but  having  a 
different  denominator.  Ans.  ^-. 

3.  A  man  has  -j-5^  of  a  barrel  of  sugar  in  one  cask,  £  in 
another,  and  |  in  another  ;  how  much  in  all  ? 

SOLUTION.  —  The  denominator  6,  of  the  second  fraction,  is  a  factor 
of  12,  the  denominator  of  the  first;  and  if  both  terms  of  f  be  multi 
plied  by  the  other  factor,  2,  it  will  become  -&.  Also  4,  the  denomi 
nator  of  the  third  fraction,  is  a  factor  of  12,  and  if  both  its  terms  be 
multiplied  by  3,  the  other  factor,  it  will  be  •£%.  And  yV  ~l~  T85~  ~h  T92 
=  ff=lJ-£  barrels.  Ans. 

4.  What  is  the  amount  of  J,  f  ,  and  f  ? 

SOLUTION.  —  As  the  denominators  are  not  factors  of  each  other,  we 
must  take  some  number  of  which  each  is  a  factor.  36  is  such  a  num 
ber.  The  first  denominator,  4,  being  a  factor  of  36,  both  terms  of 
4  may  be  multiplied  by  9,  the  other  factor,  and  we  shall  have  •$%. 
In  like  manner,  both  terms  of  f  being  multiplied  by  6,  we  have  £f  ; 
and  both  terms  of  f  being  multiplied  by  4,  we  have  §f  ;  then, 


The  process  in  the  above  examples  is  called  reducing  frac 
tions  to  a  common  denominator,  and  is  necessary  when  we 
wish  to  add  or  subtract  those  of  different  denominators.  The 
common  denominator,  it  will  be  perceived,  must  contain,  as  a 
factor,  each  of  the  other  denominators. 

It  is  not  always  manifest  what  number  will  contain  all  the 
denominators.  There  are  two  methods  of  finding  such  a 
number. 

FIRST    METHOD. 

IF  TO.  If  several  numbers  are  multiplied  together,  each 
will  evidently  be  a  factor  of  the  product.  We  have,  then,  the 
following 

Questions.  —  H  69.  How  can  eighths  and  fourths  be  added  ? 
When,  and  how,  can  one  fraction  be  reduced  to  the  denominator  of 
another?  Explain  the  third  example  ;  the  fourth.  "What  is  the  process 
called  ?  When  is  this  necessary  ?  What  must  the  common  denomina 
tor  contain  ?  What  is  not  manifest  ?  How  many  methods  of  finding  it  ? 


171.  COMMON  FRACTIONS.  89 

RULE:. 

Multiply  the  numerator  and  denominator  of  each  fraction 
by  the  product  of  the  other  denominators. 

The  several  new  denominators  will  be  products  of  the  same 
numbers,  and,  therefore,  will  be  alike  ;  and  the  numerator  and 
denominator  of  each  fraction  being  multiplied  by  the  same 
number,  its  value  is  not  altered.  See  158. 

NOTE.  — The  common  denominator  of  two  or  more  fractions  is  the 
common  multiple  of  all  their  denominators  ;  see  ^[  55. 

EXAMPLES. 

1.  Reduce  f ,  f  and  f-  to  equivalent  fractions  having  a  com 
mon  denominator. 

Each  term  of  f  being  multiplied  by  4  X  5,  or  20,  we  have  f  $. 

"         "        f     "  '*  3  X  5,  or  15,    "      "     |£. 

"  "  f  "  "  3  X  4,  or  12,  "  "  £§. 

The  terms  of  each  fraction  are  changed,  while  its  value  is  not 
altered. 

2.  Reduce  J,  f ,  J,  and  f  to  equivalent  fractions,  having  a 
common  denominator.  Ans.  £f$,  £f °>  f  j§j  4f§- 

3.  Reduce  to  equivalent  fractions,  of  a  common  denomina 
tor,  and  add  together,  J,  f,  and  J. 

A™-  *&  +  fS  +  «  =  J*  =  H&  Amount. 

4.  Add  together  |  and  ^.  Amount,  i£{L 

5.  Whatis  the  amount  of  J  +  i +|  +  |? 

Ans.  fH  =  l/TV 

6.  What  are  the  fractions  of  a  common  denominator  equiva 
lent  to  J  and  |  ?  Ans.  £f  and  f  f ,  or  T9F  and  J-J. 

!"i       -'J  SECOND    METHOD. 

^[71.  While  we  can  always  find  a  common  denominator 
by  the  above  rule,  it  will  not  always  give  us  the  least  com 
mon  denominator.  In  the  last  example,  12  as  well  as  24  is  a 
common  denominator  of  |  and  ^.  Let  us  see  how  the  12  is 
obtained. 

One  number  will  contain  another  having  several  factors, 
when  it  contains  all  these  factors.  For  example,  let  18  be 
resolved  into  the  factors  2x3x3,  which,  multiplied  togeth 
er,  will  produce  it.  It  contains  6,  the  factors  of  which,  2 
and  3,  are  the  first  and  second  factors  of  18.  It  also  contains 

Questions.  —  1  TO.  What  is  the  rule  in  the  first  method  ?  "Whence 
its  propriety  ?  What  is  a  common  multiple  ?  Explain  the  first  example 


90  COMMON  FRACTIONS.  H72. 

9,  the  fa  etc :«  of  which,  3  and  3,  are  the  second  and  third  fac 
tors  of  18.  But  it  will  not  contain  8  =  2  X  2  X  2.  for  2  is 
only  once  a  factor  in  18. 

Now  12,  the  factors  of  which  are  2  X  2  X  3,  will  contain 
4  =  2  v  2,  since  these  factors  are  the  first  and  second  factors 
of  12.  It  will  in  like  manner  contain  6.  And  it  is  the  least 
number  that  will  contain  6  and  4,  for  2  must  be  twice  a  fac 
tor,  or  it  will  not  contain  4,  and  3  must  be  a  factor,  or  it  will 
not  contain  6.  Hence,  no  one  of  these  factors  can  be  spared. 
But  24  =  2x2x2x3,  has,  it  is  seen,  2  three  times  as  a 
factor ;  so  one  2  can  be  omitted,  and  we  have  the  factors  of 
12  as  before.  We  have  2  as  a  factor  once  more  than  neces 
sary,  because  it  is  a  factor  in  both  4  and  6.  Hence,  when 
several  of  the  denominators  have  the  same  factor  we  need  retain 
it  but  once  in  the  common  denominator. 

IT  72.  The  .process  of  omitting  the  needless  factors  is 
called  getting  the  least  common  denominator  of  several  frac 
tions,  arid  is  as  follows  : 

4  and  6  are  each  divided  by  2 ;  and 
the  divisor  and  remainders  being  taken 
2.3  for  the  factors  of  the  common  denomina- 

2  X  2  X  3  =  12.      tor,  we  have  rejected  2  once. 

1.    Find  the  least  common  denominator  of  J,  f ,  §,  |,  TV 

2  .  4  .  6  .  8  .  10  SOLUTION.  — We  write  the  denomi 

nators  in  a  line,  and  divide  as  here  seen. 


1.2.3.4.     5          BV  the  first  division,  2  existing  as  a 
'  _  '  factor  in  each  of  the  five  numbers,  is 

-.,09        c          rejected    four    times,   being    retained 
*     '  once  ;  as  the  divisor  is  substituted  for 

2  X  2  X  3  X  2  X  5=  120      the  five  factors  2,  which  we  should 
have  had  by  multiplying  all  the  num 

bers  together.     But  2  being  a  factor  in  two  of  the  remainders,  it  is 
rejected  once  more  by  a  second  division. 

Ans.  120. 
2.    Find  the  least  common  denominator  of  f  ,  j7^,  and  ^f  . 


Questions.  —  ^[  71,  Why  the  necessity  of  a  second  method! 
When  will  one  number  contain  another?  What  numbers  will  18  con 
tain,  and  why?  What  will  it  not  contain?  why?  Why  will  12  contain 
the  denominators  of  both  f  and  %  1  Why  is  it  the  least  number  that  will 
contain  them  ?  Why  is  a  factor  in  24  once  more  than  necessary  ? 
What  may  then  be  done  ? 


173. 


COMMON  FRACTIONS. 


91 


FIRS 

12 
2 

12XS 

T  OPERATION.        SECO1 

8  .  12  .  24             4 

4D  OPERATION. 
8  .   12  .  24 

THI 
2 

2 
3 
2 

RD  OPERATION. 

8  .  12  .  24 

8.1.2            3 

2.3.6 

4  .    6  .  12 

4.1.1             2 
2X4  =  96,  An*. 

2.1.2 

2.3.6 

1.1.1 

2.1.2 

1.1.1 

2X2X3X2  =  24,  An* 

It  may  be  seen  that  the  product  of  the  factors  rejected  by 
the  first  operation  is  24,  while  it  is  96  by  the  second  and 
third.  The  answer  by  the  first  is  consequently  four  times 
greater,  and  is  not  the  least  common  denominator. 

Care  must  be  taken  to  avoid  this  error  in  practice.  The 
divisor  should  not  be  too  large.  It  may  always  safely,  though 
not  necessarily,  be  the  smallest  number  that  can  divide  any 
two  or  more  of  the  denominators  without  a  remainder. 

NEW   NUMERATORS. 

IT  73.  1.  Reduce  the  fractions  |,  f ,  f ,  and  f  to  equiva 
lent  fractions  having  the  least  common  denominator. 

2  2.3.4.6  SOLUTION.  — The  new  denominator  being  found, 
as  above,  to  be  12,  the  denominator,  2,  of  the  first 
fraction,  has  been  really  multiplied  by  6,  and  to 
preserve  the  equality  of  the  fraction,  the  numerator 
must  be  multiplied  by  the  same  number,  and  £  be 
comes  y8^.  So  ^  =  T82-,  |-==y92-,  and  |^  =  y42-, 
and  hence  the  fractions  are  y62,  y^-,  y9^,  ^. 

NOTE  1.  — The  factor  by  which  the  numerator  of  any  fraction  is  to 
be  multiplied,  may  be  found  by  dividing  the  common  denominator  by 
the  denominator  of  this  fraction. 

Hence,  —  For  reducing  fractions  to  their  lowest  terms. 


3 

2; 

1.3.2.3 

1.1.2.1 
<3X2  =  12 

RULE. 

Write  down  all  the  denominators  in  a  line,  and  divide  by 
the  smallest  number  greater  than  1  that  will  divide  two  or 
more  of  them  without  a  remainder.  Having  written  the  quo- 
Questions.  — 1[  72.  What  is  the  process  called  ?  In  the  first  ex 
ample,  what  factors  are  omitted,  and  what  substituted,  by  the  first  divi 
sion  ?  What  by  the  second  ?  Explain  the  second  example. 


92  COMMON  FRACTIONS.  T  73. 

tients  and  undivided  numbers  beneath,  divide  as  before ;  and 
so  on  till  there  are  no  two  numbers  which  can  be  divided  by 
a  number  greater  than  1. 

The  continued  product  of  the  quotients  and  divisors  will  be 
the  denominator  required.  Then  multiply  each  numerator 
by  the  number  by  which  its  denominator  has  really  been  mul 
tiplied. 

NOTE  2.  —  The  least  common  denominator  of  two  or  more  fractions 
is  the  least  common  multiple  of  all  their  denominators.  See  ^[  55. 

2.  Eeduce  J,  f ,  and  £  to  fractions  having  the  least  com 
mon  denominator,  and  add  them  together. 

NOTE  3. — In  writing  fractions  for  addition  and  subtraction  which 
have  a  common  denominator,  the  numerators  may  be  written  in  a  line, 
connected  by  the  appropriate  signs,  one  line  extended  under  them  all, 
and  the  denominator  written  under  this  line  but  once.  Thus,  in  the 
last  example, 

4  +  1  +  1  =  ^^^  =  ^=1^- amount. 

3.  Reduce  £  and  £  to  fractions  of  the  least  common  denom 
inator,  and  subtract  one  from  the  other. 

Aiis.  T3g-  —  -fy  =  yig-,  difference. 

4.  There  are  3  pieces  of  cloth,  one  containing  7f  yards, 
another   I3f  yards,  and  the  other   15J  yards ;  how  many 
yards  in  the  3  pieces  ? 

Before  adding,  reduce  the  fractional  parts  to  their  least  common  de 
nominator  ;  this  being  done,  we  shall  have,  — 

Adding  together  all  the  24ths,  viz.,  18  +  20  + 

7 1  =    7££  |      21,  we  obtain  59,  that  is,  ££  =  2£|.     We  write 

.3f  =  13f  £  >     down  the  fraction  ££  under  the  other  fractions,  and 

15&  =  15f^-  J      reserve  the  2  integers  to  be  carried  to  the  amount 

A  Q7i  i    ^    °f  tne  other  integers,  making  in  the  whole  3744-, 

j±ns.      £  ( g-j-  yas.  ^ng^ 

5.  There  was  a  piece  of  cloth  containing  34f  yards,  from 
which  were  taken  12§  yards ;  how  much  was  there  left  ? 

34f  =  3429¥  We  cannot  take  16  twenty-fourths,  (£},) 

12§  =  12^-f-  from  9  twenty-fourths,  (^-) ;  we  must,  there- 

.  rr~      .        fore,  borrow  1  integer,  =  24  twenty-fourths 

Ans.     4Ljfi  yds.     ^^  which?  with  ^  makeg  35  .  we  can  n(w 

Questions.  —  1f  73.  In  getting  12  as  the  common  denominator  of 
the  fractions  in  the  first  example,  by  what  number  has  the  denominator 
of  §  been  multiplied  ?  By  what,  then,  must  the  numerator  be  multiplied  ? 
The  same  questions  in  regard  to  | ;  in  regard  to  §.  Ho^.v  is  this  mul 
tiplier  found?  Give  the  rule.  What  is  the  least  common  multiple? 
What  is  done  with  the  sum  of  the  fractions  in  the  fourth  example  I 
Explain  the  borrowing  in  the  fifth  example. 


If  74,  75.  COMMON  FRACTIONS.  93 

subtract  £f  from  f  f  ,  and  there  will  remain  ££  ;  and  taking  12  inte 
gers  from  33  integers,  we  have  21  integers  remaining.     Ans. 


^T  74.     We  have,  then,  for  the  addition  and  subtraction 
of  fractions,  this  general 

RULE. 

Add  and  subtract  their  numerators  when  they  have  a  corr. 
mon  denominator  ;  otherwise,  they  must  first  be  reduced  to  a 
common  denominator. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  amount  of  f  ,  4f  ,  and  12  ?       Ans.  17£-f. 

2.  A  man  bought  a  ticket,  and  sold  f  of  it  ;  what  part  of 
the  ticket  had  he  left  ?  Ans.  f  . 

3.  Add  together  £,  |,  £,  T^,  |,  and  ££.       Amount,  2J§. 

4.  What  is  the  difference  between  14T8T  and  16^? 

Ans.  Hf 

5.  From  1J  take  |.  Remainder,  f 

6.  From  3  take  J.  Rem.  2|. 

7.  From  147J  take  48|.  Rem.  98|. 

8.  Add  together  112J,  311|,  and  lOOOf.     Ans.  1424^. 

9.  Add  together  14,  11,  4|,  T^,  and  J.  ^l?w.  30f. 

10.  From  |  take  J.     From  -£  take  |. 

11.  What  is  the  difference  between  J  and  J  ?     f  and  J? 
|  and  |?     I  and  |?     £  and  f  ?     £  and  f  ? 

12.  How  much  is  1  —  i  ?     1  —  J?     1  —  f?     1  —  |?     2 
—  |?     2—  f?    21  —  1?     3|  —  TV?     1000  —  A? 


Multiplication  of  Fractions. 

5T  75.     I.    To  multiply  a  fraction  by  a  whole  number. 

1.  If  1  yard  of  cloth  cost  J  of  a  dollar,  what  will  2  yards 
cost  ?    J  X2  =  how  much  ? 

2.  If  a  cow  consume  J  of  a  bushel  of  meal  in  1  day,  how 
much  will  she  consume  in  3  days  ?    J  X  3  =  how  much  ? 

3.  A  boy  bought  5  cakes,  at  f  of  a  dollar  each  ;  what  did 
he  give  for  the  whole  ?    f  X  5  =  how  much  ? 

4.  How  much  is  2  times  J  ? 3  times  J  ? 2  times  f  * 

Questions.  —  U  74.    Give  the  rule.    How  may  £  be  reduced  to  the 
denominator  of  |  ?  |  to  the  denominator  of  J  ?    (^[  69.) 


94  COMMON  FRACTIONS.  1175. 

5.  Multiply  ?  by  3.     f  by  2.     £  by  7. 

6.  A  woman  gives  to  her  son  |J  of  an  apple,  and  to  her 
daughter  twice  as  much ;  what  part  of  an  apple   does  the 
daughter  receive  ? 

SOLUTION.  —  She  gives  the  son  3  pieces  of  an  apple  that  had  been 
cut  into  8  pieces,  and  she  may  give  to  the  daughter  twice  the  number 
of  the  same  size,  that  is,  6  pieces,  f  X  2  =  f .  We  multiply  the 
numerator  without  changing  the  denominator. 

Or,  she  may  give  the  daughter  3  pieces  of  an  apple  that  had  been 
cut  into  half  as  many,  that  is,  4  pieces,  each  piece  being  twice  as 
large.  We  divide  the  denominator  by  2,  without  changing  the  nu 
merator,  showing  that,  as  2  small  pieces  make  1  large  piece,  the  8 
small  pieces  will  make  4  large  ones.  Ans.  f ,  or  |. 

Hence,  dividing  the  denominator,  which  is  the  divisor,  has 
the  same  effect  on  the  value  of  the  fraction  as  multiplying  the 
numerator,  which  is  the  dividend.  (IT  56.) 

Hence,  there  are  TWO  ways  to  multiply  a  fraction  by  a 
whole  number :  — 

I.  Divide  the  denominator  by  the  whole  number,  (when  it 
can  be  done  without  a  remainder,)  and  over  the  quotient  write 
the  numerator.  —  Otherwise, 

II.  Multiply  the  numerator  by  the  whole   number,  and 
under  the  product  write  the  denominator. 

If  then  the  product  be  an  improper  fraction,  it  may  be 
reduced  to  a  whole  or  mixed  number. 

EXAMPLES    FOR    PRACTICE. 

1.  If  1  man  consume  ^  of  a  barrel  of  flour  in  a  month, 

how  much  will  18  men  consume  in  the  same  time  ?  6 

men  ?  9  men  ?  Ans.  to  the  last,  1|  barrels. 

2.  What  is  the  product  of  ^  multiplied  by  40  ?     -^  X 
40  =  how  much  ?  Ans.  23§. 

3.  Multiply  y1^  by  12.     by  18.     by  21.    

oy  36.     by  48.     by  60. 

NOTE  1.  When  the  multiplier  is  a  composite  number,  we  may  first 
multiply  by  one  component  part,  and  that  product  by  the  other. 
(^[  24.)  Thus,  in  the  last  example,  the  multiplier,  60,  is  equal  to 
12  X  5;  therefore,  &  X  12  =  if,  and  |f  X  5  =  f£  =  5TV, 
Ans. 

Questions.  —  If  75.  Repeat  the  sixth  example.  Why  is  a  frac 
tion  multiplied  by  multiplying  the  numerator  ?  Why  by  dividing  the 
denominator?  Give  the  rule.  How  may  we  proceed  when  the  multi 
plier  is  a  composite  number  ?  How  is  a  mixed  number  multiplied  ? 


IF  76.  COMMON  FRACTIONS.  95 

4.  Multiply  5f  by  7. 

NOTE  2.  The  mixed  number  may  be  reduced  to  an  improper  fra,c« 
tion,  and  multiplied,  as  in  the  preceding  examples  ;  but  the  operation 
will  usually  be  shorter  to  multiply  the  fraction  and  whole  number  sep 
arately,  and  add  the  results  together.  Thus,  7  times  5  are  35  ;  and 
7  times  f  are  ^J-  =  5j,  which,  added  to  35,  make  40j,  Ans. 

Or,  we  may  multiply  the  fraction  first,  and,  writing  down  the  frac 
tion,  reserve  the  integers,  to  be  carried  to  the  product  of  the  whole 
number. 

5.  What  will  9£$  tons  of  hay  come  to,  at  17  dollars  per 
ton  ?  Ans.   lQ^\y  dollars. 

6.  If  a  man  travel  2^  miles  in  1  hour,  how  far  will  he 

travel  in  5  hours  ?  in  8  hours  ?  in  12  hours  ? 

in  3  days,  supposing  he  travel  12  hours  each  day  ? 

Ans.  to  the  last,  77f  miles. 

IT  76.     II.    To  multiply  a  whole  number  by  a  fraction. 

1.  If  36  dollars  be  paid  for  a  piece  of  cloth,  what  costs  J 
of  it? 

SOLUTION.  — If  the  price  of  1  piece  of  cloth  had  been  given  to  find 
the  price  of  several  pieces,  we  should  multiply  the  price  of  1  piece  by 
the  number  of  pieces,  and  we  must  consequently  multiply  the  price  of 
1  piece  by  the  fraction  of  a  piece  where  the  price  of  a  fraction  is  re 
quired. 

The  price  of  1  piece,  36  dollars,  must  be  multiplied  by  |.  One 
fourth  of  the  cloth  would  cost  £  of  36,  or  9  dollars,  and  I  would  cost 
3  times  as  much,  9  X  3=27.  Ans.  27  dollars. 

The  product  is  f  of  the  multiplicand,  a  part  denoted  by  the 
multiplying  fraction. 

Multiplication,  therefore,  when  applied  to  fractions,  does 
not  always  imply  increase,  as  in  whole  numbers ;  for,  when 
the  multiplier  is  less  than  unity,  it  will  always  require  the 
product  to  be  less  than  the  multiplicand,  to  which  it  would  be 
equal  if  the  multiplier  were  1. 

There  are  two  operations,  a  division  and  a  multiplication. 
But  it  is  matter  of  indifference,  as  it  respects  the  result,  which 
of  these  operations  precedes  the  other,  for  36  X  3  -5-  4  =  27, 
the  same  as  36  -j-  4  X  3  =  27. 

Hence,   To  multiply  by  a  fraction,  we  have  this 

RULE. 

Divide  the  multiplicand  by  the  denominator  of  the  multi- 
Questions.  —  Tf  76.    WnY  must  36  be  multiplied  by  |  ?  How  does 
the  product  compare  with  the  multiplicand,  and  why  ?    Give  the  rale. 


96  COMMON  FRACTIONS.  IF  77,  78. 

plying  fraction,  and  multiply  the  quotient  by  the  numerator  ; 
or,  when  there  would  be  a  remainder  by  division,  first  multi 
ply  by  the  numerator,  and  divide  the  product  by  the  denomi 
nator. 

2.  What  is  the  product  of  90  multiplied  by  J  ? 

Ans.  45. 

3.  Multiply  369  by  §. 

4.  Multiply  45  by  TV  Product,  31  1. 

5.  Multiply  210  by  |. 

6.  Multiply  1326  by  -ft.  Prod.  241^. 


NOTE.  —  As  either  factor  may  be  the  multiplier,  (T[21,)  we  may 
multiply  by  the  whole  number,  making  the  fraction  the  multiplicand. 
Hence,  the  examples  in  this  and  ^]  75,  may  be  performed  by  the  same 
rule. 

IT  77.  1.  At  40  dollars  for  1  acre  of  land,  what  will  $• 
of  an  acre  cost  ?  40  X  f  =  how  much  ? 

In  this  example,  the  price  of  1  acre,  40  dollars,  is  multiplied 
by  the  fraction  of  an  acre,  f  .  Ans.  32  dollars. 

Hence,  When  the  price  of  unity  is  given,  to  find  the  cost 
of  any  quantity,  less  or  more  than  unity, 

RULE. 

Multiply  the  price  by  the  quantity. 

EXAMPLES    FOR    PRACTICE. 

2.  If  a  ship  be  worth  1367  dollars,  what  is  f  of  it  worth  ? 

Ans.  303£  dollars. 

3.  What  cost  |£  °f  a  ton  °f  butter,  at  225  dollars  per  ton  ? 

Ans.  190&  dollars. 

IF  78»     III.    To  multiply  one  fraction  by  another. 

1.  At  f  of  a  dollar  for  one  bushel  of  corn,  what  will  f  of  a 
bushel  cost  ?  £  X  f  =  how  much  ? 

SOLUTION.  —  The  price  of  one  bushel,  £,  is  to  be  multiplied  by 
the  fraction  of  a  bushel,  |,  (^[  77.) 

We  first  divide  f-  by  3,  to  get  the  price  of  J  of  a  bushel.  This 
we  can  do  by  multiplying  the  denominator  by  3,  thus  making  the 
parts  of  a  dollar  only  one  third  as  large,  (15ths,)  while  the  same 
number,  4,  is  taken.  £-7-3  =  ^  of  a  dollar,  the  price  of  one 

Questions.  —  If  77.  Explain  the  first  example.  What  two  things 
are  given,  and  what  required  ?  Eule. 


IT  79.  COMMON  FRACTIONS.  97 

third  ;  and  ^  X  2  =  •&  of  a  dollar,  price  of  f  of  a  bushel.     Ans. 
j§g-  of  a  dollar. 

The  denominator  5  of  the  multiplicand  is  multiplied  by  3, 
the  denominator  of  the  multiplier,  and  4,  the  numerator  of  the 
multiplicand,  by  2,  the  numerator  of  the  multiplier. 

Hence,  To  multiply  one  fraction  by  another, 

RULE. 

Multiply  the  denominators  together  for  the  denominator  of 
the  product,  and  the  numerators  for  the  numerator  of  the 
product. 

By  this  process  the  multiplicand  is  divided  by  the  denomi 
nator  of  the  multiplying  fraction,  and  the  quotient  multiplied 
by  the  numerator,  as  in  11  76. 

EXAMPLES  FOR  PRACTICE. 

2.  Multiply  &  by  f     Multiply  ^  by  f  .        Product,  ^. 

3.  At  j-fg  of  a  dollar  a  yard,  what  will  £  of  a  yard  of  cloth 
cost  ? 

4.  At  6|  dollars  per  barrel  for  flour,  what  will  T7^  of  a  bar 
rel  cost  ? 

NOTE.  —  Mixed  numbers  must  be  reduced  to  improper  fractions. 

6§  =  -V-  ;  then  fy  X  TV  =  f  f  &  =  2T^F  dollars,  Ans. 

5.  At  f  of  a  dollar  per  yard,  what  cost  7|  yards  ? 

Ans.  6J^  dollars. 

6.  At  2J  dollars  per  yard,  what  cost  6f  yards  ? 

Ans.  14§f  dollars. 

IT  79.  1.  What  will  §  of  a  yard  of  cloth  cost  at  £  of  a 
dollar  per  yard  ? 

SOLUTION.  —  We  multiply  the  price  of  1  yard,  §,  by  f,  the  frac 
tion  of  a  yard,  (^[78.)  Getting  the  price  of  |  of  a  yard  is  getting 
|  of  f  of  a  dollar.  |  of  f  is  an  expression  called  a  compound  frac 
tion,  ("If  65.)  The  reducing  of  a  compound  fraction  to  a  simple  one 
is,  then,  the  same  as  multiplying  fractions  together. 

2.  What  is  $  of  J?  f  ofTV  J  of  |  ? 

3.  How  much  is  §  of  f  of  f  ? 

§  of  f  we  have  found  to  be  -j8^,  and  ^  of  f  by  the  above 
rule  is  -,  Ans.  Hence, 


Questions.  —  1T  78.     What  is  the  first  operation,  Ex.  1  ?    Whence 
its  propriety?    Second  operation?    Rule.    What  is  done  by  the  first 
operation  required  in  the  rule  ?  by  the  second  operation  ? 
9 


98  COMMON  FRACTIONS.  IT  80 

The  word  of  between  fractions  implies  their  continued  mul 
tiplication.  If  there  are  more  than  two  fractions,  we  multiply 
together  the  several  numerators  and  the  several  denominators. 

4.  How  much  is  &  of  f  of  f  of  f  ?      Ans.  •££&  ==  yfe. 

5.  How  much  is  f  of  f  of  £  of  f  ? 


If  8O.     1.    How  much  is  ^  of  f  of  f  of  f  ? 

Since  the  numerators  are  to  be  multiplied  together,  and 
their  product  to  be  divided  by  the  product  of  the  denominators, 
The  operation  may  be  shortened  by  Cancelation,  (IF  60.) 

OPERATION.  SOLUTION.  —  Performing  this   opera- 

$  tion,  as  described  in  •fl  60,  we  have  can- 

9         $        $        2        1      celed  all  the  factors  of  the  numerators, 

97*  °f  t±  °f  ~4  °f  «5  ==  A'     anc^  have  the  factors  2,  2,  remaining  of 

043^      the  denominators.      But  the  numerator 

22%  2  =  2  X  1,  the  numerator  5  =  5  X  1,  3 

=  3X  ls  &c.,  and  we  have  in  reality 

the  fectors  1,  1,  1,  and  1,  left  in  the  numerators.  1X1X1X1  =  1 
for  the  new  numerator,  and  2X2X1X1  =  4  for  the  new  denomina 
tor.  Hence,  when  all  the  factors  excepting  the  1's  in  the  numerators 
or  denominators  cancel,  the  new  numerator  or  denominator,  as  the  case 
may  be,  will  be  1. 

EXAMPLES  FOR  PRACTICE  IX  CAXCELATIOX. 

2.  |  of  |  of  f  of  |  of  -&  of  £  of  f  =how  much  ?  Ans.  •&. 

3.  What  is  the  continual  product  of  7,  J,  £  of  f  and  3£  ? 

NOTE.  —  The  integer  7  may  be  reduced  to  the  form  of  an  improper 
fraction,  by  writing  a  unit  under  it  for  a  denominator,  thus,  -£. 

Ans.  2|£. 

4.  What  is  the  continued  product  of  3,  f  ,  f  of  |,  2^-,  and 
y  off  off?  Anx.3%%. 

5.  Reduce  f  of  £  of  f  of  £  of  22£  to  a  simple  fraction. 

Ans.  9£$. 

6.  A  horse  consumed  f  of  J  of  8  tons  of  hay  in  one  win 
ter  ;  how  many  tons  did  he  consume  ?  ^715.  2f-  tons. 

7    Reduce  £  of  j  of  f  of  f  of  f  of  1  to  a  simple  fraction. 

Ans.  -. 


Questions.  —  If  79.  How  does  it  appear  that  we  have  a  compound 
fraction  in  the  first  example  ?  What  does  the  word  of  between  fractions 
imply  ?  What  ;.s  done  when  there  are  more  than  two  fractions  ? 

If  80.  Why  can  cancelation  be  applied  to  the  multiplication  of  frac 
tions  ?  Explain  the  pro  ;ess.  What  is  done  with  integers  when  occurring 
with  fractions  ? 


H81.  COMMON  FRACTIONS.  99 

1T8O.    (2.)     PROMISCUOUS  EXAMPLES  IX  THE  MULTI 
PLICATION    OF  FRACTIONS. 

1.  At  |  dollars  per  yard,  what  cost  4  yards  of  cloth  ?  

5  yards  ?  6  yards  ?  8  yards  ?  20  yards  ? 

Am.  to  the  last,  15  dollars. 

2.  Multiply  148  by  J.  by  £.  by  &.  by  T30. 

Last  product,  44jV 

3.  If  2^  tons  of  hay  keep  1  horse  through  the  winter, 

how  much  will  it  take  to  keep  3  horses  the  same  time  ?  

7  horses  ?  13  horses  ?  Ans.  to  last,  37^  tons. 

4.  What  will  8^-  barrels  of  cider  come  to,  at  3  dollars  per 
barrel  ?  Am.  25J  dollars. 

5.  At  14|  dollars  per  cwt.,  what  will  be  the  cost  of  147 
cwt.  ?  Am.  2168J  dollars. 

6.  A  owned  f  of  a  ticket ;    B  owned  T\  of  the  same ;  the 
ticket  was  so  lucky  as  to  draw  a  prize  of  1000  dollars  ;  what 
was  each  one's  share  of  the  money  ? 

Ans.  A's  share,  600  dollars ;  B's  share,  400  dollars. 

7.  Multiply  J  of  |  by  j  of  fc.  Product,     |. 

8.  Multiply  7£  by  2TV  "       15J. 

9.  Multiply  I  by  2|.  "         2J. 

10.  Multiply  |  of  6  by  f.  "  1. 

11.  Multiply  |  of  2  by  \  of  4.  «  3. 

12.  Multiply  continually  together  £  of  8,  f  of  7,  f  of  9,  and 
J  of  10.  Product,  20. 

7  13.   Multiply  1000000  by  f .  Product,  555555%. 


Division  of  Fractions. 

IT  81.  I.    To  divide  a  fraction  by  a  whole  number. 

1.  If  2  yards  of  cloth  cost  |  of  a  dollar,  what  does  1  yard 
cost  ?  how  much  is  |  divided  by  2  ? 

2.  If  a  cow  consume  f  of  a  bushel  of  meal  in  3  days,  how 
much  is  that  per  day  ?    f  -f-  3  =  how  much  ? 

3.  If  a  boy  divide  f  of  an  orange  among  2  boys,  how  much 
will  he  give  each  one  ?     f  -5-  2  =  how  much  ? 

4.  A  boy  bought  5  cakes  for  •££  of  a  dollar;  what  did  1 
cake  cost  ?    •}§•  - .-  5  =  how  much  ? 

5.  If  2  bushels  of  apples  cost  ^  of  a  dollar,  what  is  that  per 
bushel  ? 

1  bushel  is  the  half  of  2  bushels  ;  the  half  of  f  is  £. 

Ans.     dollar 


100  COMMON  FRACTIONS.  If  81. 

6.  If  3  horses  consume  -ff  of  a  ton  of  hay  in  a  month,  what 
will  1  horse  consume  in  the  same  time  ? 

SOLUTION. — ^-§  are  12  parts  ;  if  3  horses  consume  12  such  parts 
in  a  month,  as  many  times  as  3  are  contained  in  12,  so  many  parts  1 
horse  will  consume.  Ans.  -fe  of  a  ton. 

Hence,  we  divide  a  fraction  by  dividing  the  numerator  with 
out  changing  the  denominator,  taking  a  less  number  of  parts 
of  the  same  size. 

7.  A  woman  would  divide  f  of  a  pie  equally  between  her 
two  children  ;  how  much  does  each  receive  ? 

SOLUTION.  —  She  cannot  divide  the  3  pieces  into  2  equal  parts  and 
leave  them  all  whole.  But  as  the  denominator  4  shows  into  how 
many  parts  the  pie  is  cut,  multiplying  it  by  2  is  equivalent  to  cutting 
the  pie  into  twice  as  many,  or  8  pieces  of  half  the  size.  That  is,  we 
may  cut  each  piece  into  2  equal  parts,  and  give  1  of  them  to  each 
child,  who  will  then  have  the  same  number  of  pieces,  3,  only  half  as 
large.  Ans.  f . 

Hence,  a  fraction  is  divided  by  multiplying  its  denominator 
ivithout  changing  its  numerator,  as  the  parts  are  made  small 
er,  while  the  same  number  is  taken. 

Multiplying  the  denominator,  then,  which  is  the  divisor, 
has  the  same  effect  on  the  fraction  as  dividing  the  numerator, 
which  is  the  dividend,  (IF  57.) 

NOTE  1.  —  By  comparing  this,  and  ^[  75,  we  shall  see  that  where 
either  term  of  a  fraction  is  to  be  multiplied  or  divided,  the  contrary 
operation  may  be  performed  on  the  other  term. 

Hence,  we  have  TWO  ways  to  divide  a  fraction  by  a  whole 
number :  — 

I.  Divide  the  numerator  by  the  whole  number,  (if  it  will 
contain  it  without  a  remainder,)  and  under  the  quotient  write 
the  denominator.  — Otherwise, 

II.  Multiply  the  denominator  by  the  whole  number,  and 
over  the  product  write  the  numerator. 


Questions.  —  IT  81.  How  are  -J-f  divided  by  4  ?  What  difficulty 
in  dividing  3  pieces  of  pie  among  2  children  ?  How  then  may  |  be  di 
vided?  Why  does  multiply! ig  the  denominator  divide  the  fraction? 
In  what  two  ways  is  a  fractio  a  divided  ?  Apply  ^[  57  to  the  operation. 
What  appears  from  comparing  this  with  Tf  75  ?  Repeat  the  rule.  How 
do  you  divide  a  fraction  by  a  composite  number  ?  How  divide  a  mixed 
number  ?  How,  when  the  mixed  number  is  large  ? 


U82.  COMMON  FRACTIONS.  101 

EXAMPLES  FOR  PRACTICE. 

8.  If  7  pounds  of  coffee  cost  f  ^  of  a  dollar,  what  is  that  per 
pound  ?  f£  -r-  7  =  how  much  ?  Ans.  /-%  of  a  dollar. 

9.  If  £|{  of  an  acre  produce  24  bushels,  what  part  of  an 
acre  will  produce  1  bushel  ?  £§  -5-  24==  how  much  ? 

10.  If  12  skeins  of  silk  cost  -f£  of  a  dollar,  what  is  that  a 
skein  ?    ^-^  -;-  12  =  how  much  ? 

11.  Divide  f  by  16. 

NOTE  2.  —  When  the  divisor  is  a  composite  number,  we  can  first 
divide  by  one  component  part,  and  the  quotient  thence  arising  by  the 
other,  (Tf  39.)  Thus,  in  the  last  example,  16  =  8  X  2,  and  f  -4- 
8  =  £,  and  £  -T-  2  =  ^.  Ans.  ^ . 

12.  Divide  &  by  12.    Divide  -fa  by  21.    Divide  f  f  by  24. 

13.  If  6  bushels  of  wheat  cost  4J  dollars,  what  is  it  per 
bushel  ? 

NOTE  3.  —  The  mixed  number  may  be  reduced  to  an  improper  frac 
tion,  and  divided  as  before. 

Ans.  ^f-  =  ^  of  a  dollar,  expressing  the  fraction  in  its  lowest 
terms. 

14.  Divide  4f£  dollars  by  9.  .  .  Quot.  -/3  qf  a  dollar. 

15.  Divide  12f  by  5.           .  Qu'J(:\8-  =  2$. 

16.  Divide  14|  by  8.  Quot.  IfjL 

17.  Divide  184$  by  7.  0:.^  '.  '47/^616^. 

NOTE  4.  —  When  the  mixed  number  is  large,  it  will  be  most  conve 
nient,  first,  to  divide  the  whole  number,  and  then  reduce  the  remainder 
to  an  improper  fraction  ;  and,  after  dividing,  annex  the  quotient  of  the 
fraction  to  the  quotient  of  the  whole  number ;  thus,  in  the  last  ex 
ample,  dividing  184£  by  7,  as  in  whole  numbers,  we  obtain  26  integers 
with  2j=|-  remainder,  and,  dividing  this  by  7,  we  have  T5¥,  and 
264-^  =  26-^.  Ans. 

18.  Divide  2786J  by  6.  Ans.  464|. 

19.  How  many  times  is  24  contained  in  7646££  ? 

Ans. 

20.  How  many  times  is  3  contained  in  462|  ? 

Ans. 

IF  82.     I7 .    To  divide  a  whole  number  by  a  fraction. 

1.  A  man  would  divide  9  dollars  among  some  poor  persons, 
giving  them  j  of  a  dollar  each ;  how  many  will  receive  the 
money  ? 

q* 


102  COMMON  FRACTIONS. 

SOLUTION.  —  We  wish  to  see  how  many  times  $  of  a  dollar  is  con 
tained  in  9  dollars.     But 

OPERATION.  as  the  divisor  is  4th8,  (25 

9  cent  pieces,)  we  must  re- 

4  duce  *he  dividend  to  4llls, 

as  both  must  be  of  the 

4:tht  of  a  dollar,  3)36  ^  of  a  dollar,      same    denomination,    (^[ 

33  ;)  thus,  we  multiply  9 

12  persons.  by  4,  to  reduce  it  to  4th3, 

since  there  are  4  fourths 

in  one  dollar.     Then,  as  many  times  as  3  fourths  are  contained  in  36 
fourths,  so  many  persons  will  receive  the  money. 

We  find  the  number  to  be  12  persons,  a  number  greater  than  the 
dividend  or  number  of  dollars.  Division,  then,  when  applied  to  frac 
tions,  does  not  always  imply  decrease.  The  quotient  is  greater  than 
the  dividend  when  the  divisor  is  less  than  1,  to  which  it  is  just  equal 
when  the  divisor  is  1. 

Hence,  To  divide  a  whole  number  by  a  fraction, 

RULE. 

Multiply  the  dividend  by  the  denominator  of  the  dividing 
fraction,  (thereby  reducing  the  dividend  to  parts  of  the  same 
magnitude  as  the  divisor,)  and  divide  the  product  by  the  nu 
merator.  t  •...,* 

EXAM£-L,KS  FOR   PRACTICE. 

2.  Jiew  ir.rvny  time's  iM  J  contained  in  7  ?     7  -5-  \  =  how 
many?  -' '>*•,'••'' 

3.  How  many  times  can  I  draw  J  of  a  gallon  of  wine  out 
of  a  cask  containing  26  gallons  ? 

4.  Divide  3  by  }.     6  by  |.     10  by  f . 

5.  If  a  man  drink  T9^  of  a  quart  of  beer  a  day,  how  long 
will  3  gallons  last  him?  Anx.  211  days. 

6.  If  2|  bushels  of  oats  sow  an  acre,  how  many  acres  will 
22  bushels. sow  ?     22  -r  2f  =  how  many  times  ? 

NOTE.  —  Reduce  the  mixed  number  to  an  improper  fraction,  2| 
=Jji-.  Ans.  S  acres 

7.  How  many  times  is  -^  contained  in  6  ? 

Ans.  f  of  1  time. 

8.  How  many  times  is  Sf-  contained  in  53  ? 

Ans.  6|f  times. 

Questions.  —  ^[  82.  How  is  the  principle  that  the  divisor  and  divi 
dend  mas1,  be  of  the  same  denomination  applied  to  the  first  example? 
When  is  the  quotient  greater  than  the  dividend,  and  when  equal  to  it  ? 
Give  the  rule  for  dividing  a  whole  number  by  a  fraction. 


H  83,  84.  COMMON  FRACTIONS.  103 

IT  83.  1.  At  |  of  a  dollar  per  yard,  how  much  cloth  can 
be  bought  for  12  dollars  ? 

SOLUTION.  —  As  many  times  as  f  of  a  dollar  is  contained  in  (or  can 
be  subtracted  from)  12  dollars,  so  many  yards  can  be  bought. 

Ans.  18  yards. 

Hence,  When  the  price  of  unity  and  the  price  of  any  quan 
tity  are  given,  to  find  the  quantity, 

RULE. 

Divide  the  price  of  the  quantity  by  the  price  of  unity. 

EXAMPLES  FOR  PRACTICE. 

2.  At  4f  dollars  a  yard,  how  many  yards  of  cloth  may  be 
bought  for  37  dollars  ?  Ans.  S^  yards. 

3.  At  Y9^  of  a  "dollar  a"  pound,  how  many  pounds  of  tea 
may  be  bought  for  84  dollars  ?  Ans.  90J  pounds. 

4.  At  |-  of  a  dollar  for  building  1  rod  of  stone  wall,  how 
many  rods  may  be  built  for  87  dollars  ?     87  -f-  f-  =  how  many 
times  ?  Ans.  104f  rods. 

IT  84.     III.    To  divide  one  fraction  by  another. 

1.  At  |  of  a  dollar  per  bushel,  how  many  bushels  of  oats 
can  be  bought  for  £  of  a  dollar  ? 

SOLUTION,  -r-  We  are  to  divide  £  by  f.  Hf83.)  To  divide  by  a 
fraction  we  multiply  the  dividend  by  the  denominator  of  the  dividing 
fraction,  and  divide  the  product  by  the  numerator,  (^f  82.) 

£  X9=-VS  and  -^5 -j-2=^=3|  bushels,  Ans. 

Hence, 

RULE. 

Invert  the  divisor,  and  multiply  together  the  two  upper 
terms  for  a  numerator,  and  the  two  lower  terms  for  a  denomi 
nator; — for  thereby  the  numerator  of  the  dividend  is  multi 
plied  by  the  denominator  of  the  divisor,  and  thus  the  dividend 
is  multiplied  by  this  denominator,  and  the  denominator  of  the 
dividend  is  multiplied  by  the  numerator  of  the  divisor,  and 
thus  the  dividend  is  divided  by  this  numerator,  as  in  IF  82. 


Questions.  —  ^[  83.  What  two  things  are  given,  and  what  required, 
Ex.1?  Rule. 

IF 84.  How  do  we  divide  by  a  fraction?  How  multiply  a  fraction? 
How  divide  a  fraction?  Rule  for  dividing  one  fraction  by  another. 
What  thereby  is  done  ? 


104  COMMON   FRACTIONS.  If  85. 

EXAMPLES  FOR  PRACTICE. 

2.    Divide  J  by  J.      Quot.  1.     Divide  J  by  |.         Qwo£.  2. 


3.  Divide     by    .     Quot.  3.     Divide     by  TV     Qzwtf.  f  £. 

4.  If  4f  pounds  of  butter  serve  a  family  1  week,  how  many 
weeks  will  36£  pounds  serve  them  ? 

NOTE.  —  The  mixed  numbers,  it  will  be  recollected,  may  be  reduced 
to  improper  fractions.  Ans.  8Tf¥  weeks. 

5.  Divide  2J  by  1  J.  Divide  lOf  by  2|. 

Quot.  1|. 

6.  How  many  times  is  ^  contained  in  f  ? 

^.715.  4  times. 

7.  How  many  times  is  f  contained  in  4£  ? 

,4?w.  113  times. 

8.  Divide  |  of  |  by  J  of  f  Qwo*.  4. 

IT  85.  1.  If  £  of  a  yard  of  cloth  cost  f  of  a  dollar,  what 
is  that  per  yard  ? 

SOLUTION.  —  Had  the  price  of  several  yards  been  given,  we  would 
divide  it  by  the  number  of  yards,  to  find  the  price  of  1  yard,  and,  in 
like  manner,  we  must  divide  the  price  of  the  fraction  of  a  yard  (f 
of  a  dollar)  by  the  fraction  of  a  yard,  (£  ,)  to  find  the  price  of  1  yard. 

Ans.  §£  of  a  dollar  per  yard. 

Hence,  When  the  price  of  any  quantity  less  or  more  than 
unity  is  given,  to  find  the  price  of  unity, 

RULE. 

Divide  the  price  by  the  quantity. 

EXAMPLES  FOR  PRACTICE. 

2.  At  £  of  a  dollar  for  3J  bushels  of  apples,  what  does  1 
bushel  cost  ?  Ans.  J  of  a  dollar. 

3.  At  £  of  a  dollar  for  4|  bushels  of  oats,  what  does  I 
bushel  cost  ?  Ans.  -fa  of  a  dollar. 

IT  85.  (2.)     Reduction  of  complex  to  simple  fractions. 
1.   What  simple  fraction  is  equivalent  to  the  complex  frac 
tion  |? 

SOLUTION.  —  Since  the  numerator  of  a  fraction  is  a  dividend  of 

Questions.  —  If  85.  What  two  things  are  given,  and  what  required, 
Ex.  1  ?  Give  the  rule. 


1Tb5.  COMMON  FRACTIONS.  105 

which  the  denominator  is  the  divisor,  we  may  divide  |  by  £,  by  the 
rule,  1 84.      |  -r-  f  =  |£.  Ans. 

42 

2.  What  simple  fraction  is  equal  to  -=•  ? 

OPERATION.  SOLUTION.  —  We  reduce  4|  to  the  im- 

4|  =  -^,  #?id  proper  fraction  -^,  which  we  divide  by  7, 

Jj§-  -7-  7  =  £!•,  .Aws.      according  to  the  rule,  ^[  81. 

The  above  illustrations  are  sufficient  to  establish  the  fol 
lowing 

RULE. 

Reduce  any  mixed  number  which  may  occur  in  the  com 
plex  fraction  to  the  fractional  form,  or  any  compound  fraction 
to  a  simple  one,  after  which  divide  the  numerator  by  the  de 
nominator,  according  to  the  ordinary  rules  for  the  division  of 
fractions. 

EXAMPLES  FOR  PRACTICE. 

7-i 

3.  Reduce  the  complex  fraction  -f-  to  a  simple  one. 


4.  What  is  the  value  of  -2  ?  Ans.  19. 

5.  What  simple  fraction  is  equal  to  ~  ?  Ans.  || 

y 

6.  What  simple  fraction  is  equal  to  J|  ?  Ans.  gfo 

7.  What  simple  fraction  is  equal  to  -~  ?          Ans.  f  f f- 

8.  What  simple  fraction  is  equal  to  —  ?  Ans.  A^= 

o 

9.  What  simple  fraction  is  equal  to  —^  ?  ^TZS. 

10.  What  simple  fraction  is  equal  to  —  ?  Ans. 

11.  What  simple  fraction  is  equal  to  \  7  5         ? 

X>  ^3 


Questions.  —  If  85.  (2.)  How  is  the  complex  fraction,  Ex.  1,  re 
duced  to  a  simple  one  ?  wny  ?  Give  the  nil*'  for  reducing  complex  to 
simple  fractions. 


106  COMMON  FRACTIONS.  IF  85,86 

1F  85.  (3.)  PROMISCUOUS  EXAMPLES  ix  THE  DIVI 

SION    OF    FRACTIONS. 

1.  If  7  Ib.  of  sugar  cost  -^  of  a  dollar,  what  is  it  per 
pound  ?     £fa  -5-  7  =  how  much  ?     ^  of  T6^  is  how  much  ? 

2.  At  £  of  a  dollar  for  f  of  a  barrel  of  cider,  what  is  that 
per  barrel  ?  Ans.  -3-  of  a  dollar. 

3.  If  4  pounds  of  tobacco  cost  £  of  a  dollar,  what  does  1 
pound  cost  ?  Ans.  -fa  doll. 

4.  If  J  of  a  yard  cost  4  dollars,  what  is  the  price  per  yard  ? 

Ans.  4f  dollars. 

5.  If  14£  yards  cost  75  dollars,  what  is  the  price  per  yard? 

Ans.  5-^  dollars. 

6.  At  31  \  dollars  for  10J  barrels  of  cider,  what  is  that  per 
carrel  ?  Ans.  3  dollars. 

7.  How  many  times  is  f  contained  in  746  ?   Ans.  19S9|. 

8.  Divide  J  of  |  by  j.  Divide  £  by  f  of  f  . 

Qwcrf.  f  .  Quot.  3£f 

9.  Divide  J  of  £  by  £  of  |.  QwoZ.  Jf  . 

10.  Divide  |  of  4  by  ^.  QMO*.  3. 

11.  Divide  4f  by  $  of  4.  Qz^tf.  2^ 

12.  Divide  $  of  4  by  4f  .  Qw<tf  .  ff, 


13.    Divide        by  Quot.  95}  . 

y         7 


^  86.    Review  of  Common  Fractions. 

Questions.  —  "What  are  fractions  ?  Whence  is  it  that  the  parts  into 
which  any  thing  or  any  number  may  be  divided,  take  their  name? 
What  determines  the  magnitude  of  the  parts  ?  Why  ?  How  does  increas 
ing  the  denominator  affect  the  value  of  the  fraction?  Increasing  the 
numerator  affects  it  how  ?  How  is  an  improper  fraction  reduced  to  a 
whole  or  mixed  number  ?  How  is  a  mixed  number  reduced  to  an  im 
proper  fraction  ?  a  whole  number  ?  How  is  a  fraction  reduced  to  its 
most  simple  or  lowest  terms?  How  is  a  common  divisor  found?  (^f  61.) 
the  greatest  common  divisor?  (^[  62.)  Whence  the  necessity  of  reduc 
ing  fractions  to  a  common  denominator  ?  When  may  one  fraction  be 
reduced  to  the  denominator  of  another?  What  must  the  common 
denominator  be?  (If  69.)  Give  the  first  method  of  finding  it,  and 
the  principles  on  which  it  is  founded ;  the  second  method,  and  the  prin 
ciples.  What  is  understood  by  a  multiple  ?  by  a  common  multiple  ?  by 
the  least  common  multiple?  What  is  the  process  of  finding  it?  ($  72.) 
How  are  fractions  added  and  subtracted  ?  How  many  ways  are  there 
to  multiply  a  fraction  by  a  whole  number?  How  does  it  appear,  that 
dividing  the  denominator  multiplies  the  fraction?  How  is  a  mixed  number 
multiplied  ?  What  is  implied  in  multiplying  by  a  fraction  ?  Of  what 
operations  does  it  consist  ?  When  the  multiplier  is  less  than  a  unit,  what 
is  the  prod  a^t  compared  with  the  multiplicand  ?  What  two  things  are . 


1TS6.  COMMON  FRACTIONS.  107 

given,  and  what  required  in  f  77  ?  What  in  ^[  83  ?  What  in  ^f  85  ? 
Explain  the  principle  of  multiplying  one  fraction  by  another.  Of  divid 
ing  one  fraction  by  another.  How  do  you  multiply  a  mixed  number 
by  a  mixed  number  ?  How  does  it  appear,  that  in  multiplying  both 
terms  of  the  fraction  by  the  same  number,  the  value  of  the  fraction  is 
not  altered?  How  many,  and  what  are  the  ways  of  dividing  a  fraction 
by  a  whole  number  ?  How  does  it  appear  that  a  fraction  is  divided  by 
multiplying  its  denominator  ?  How  does  dividing  by  a  fraction  differ  from 
multiplying  by  a  fraction  ?  When  the  divisor  is  less  than  a  unit,  what  is 
the  quotient  compared  with  the  dividend  ?  How  do  you  divide  a  whole 
number  by  a  fraction  ? 

EXERCISES. 

•tj 

1.  What  is  the  amount  of  £  and  f  ?  -  of  J  and  f  ?  -- 
of  12J,  3|,  and  4|  ?  Am.  to  the  last,  20|£. 

2.  How  much  is  J  less  J  ?     ^  —  £?     -&  —  ^  ?     14J  — 
4f?     6-4f?     «»-Jof§ofj? 

Ans.  to  the  last,  £f£. 

3.  What  fraction  is  that,  to  which  if  you  add  f  the  sum 
will  be  $?  Ans.  tf. 

4.  What  number  is  that,  from  which  if  you  take  f  the 
remainder  will  be  £  ?  -4?zs.  f  f  . 

5.  What  number  is  that,  which  being  divided  by  j  the 
quotient  will  be  21  ?  Ans.  15|. 

6.  What  number  is  that,  which  multiplied  by  f  produces  J  ? 


7.  What  number  is  that,  from  which  if  you  take  f  of  itself 
the  remainder  will  be  12  ?  Ans.  20. 

8.  What  number  is  that,  to  which  if  you  add  §  of  f  of 
itself  the  whole  will  be  20  ?  Ans.  12. 

9.  What  number  is  that,  of  which  9  is  the  §  part.  ? 

Ans.  13J. 

10.  At  f  of  a  dollar  per  yard,  what  costs  f  of  a  yard  of 
cloth  ?  Ans.  ££  of  a  dollar. 

11.  At  5f  dollars  per  barrel,  what  costs   18^  barrels  of 
flour  ?  Ans.  108^  dollars. 

12.  What  costs  84  pounds  of  cheese,  at  -fa  of  a  dollar  per 
pound?  Ans.  11^  dollars. 

13.  What  cost  45  yards  of  gingham,  at  f  of  a  dollar  per 
yard  ?  Ans.  28  £  dollars. 

14.  What  must  be  paid  for  -j7^  of  a  yard  of  velvet,  at  5  dol 
lars  per  yard  ?  Ans.  2T3F  dollars. 

15.  If  |-  of  a  pound  of  tea  cost  -^  of  a  dollar,  what  is  that 
per  p<;  und  ?  Ans.  -j^  of  a  dollar. 

16.  If  7£  barrels  of  pork  cost  73£  dollars,  what  is  that  per 
oarrel  ?  Ans.  10£  dollars. 


JOS  DECIMAL   FRACTIONS.  f  87. 

tl7.  If  4  acres  of  land  cost  82-j^  dollars,  what  is  that  pe-r 
acre  ?  Ans.  20££  dollars. 

18.  At  -£%  of  a  dollar  for  3£  bushels  of  lime,  what  costs  1  . 
bushel  ?  A?is.  -fa  of  a  dollar. 

19.  Paid  4f  dollars  for  coffee,  at  ^  of  a  dollar  per  pound  , 
how  many  pounds  did  I  buy  ?  Ans.  29£  pounds. 

20.  At  If  dollars  per  bushel,  how  much  wheat  may  be 
bought  for  82|  dollars  ?  Ans.  59^  bushels. 

21.  If  8|  yards  of  silk  make  a  dress,  and  9  dresses  be 
made  from  a  piece  containing  80  yards,  what  will  be  the  rem 
nant  left  ?  Ans.  1  J  yards. 

NOTE.  —  Let  the  pupil  reverse  and  prove  this,  and  the  following 
example. 

22.  How  many  vests,  containing  £•  of  a  yard  each,  can  be 
made  from  22  yards  of  vesting?  what  remnant  will  be  left? 

Ans.  25  vests.     Remnant,  £  yard. 

23.  What  number  is  that,  which  being  multiplied  by  15 
the  product  will  be  }  ?  Ans.  %*$. 

24.  What  is  the  product  of  %  °^  into  £-  ? 

7  &fj 

Ans.  fW 

25.  Which  of  the  eleven  numbers,  8,  9,  J  1,  12,  14,  15,  16, 
.  8,  20,  22,  24,  have  all  their  factors  the   same  as  factors  in 
72? 

NOTE.  —  The  72  must  be  resolved  into  the  greatest  number  of  fac- 
lors  possible,  which  are  2,  2,  2,  3,  3.  In  like  manner,  each  of  the 
other  numbers  must  be  resolved.  Ans.  8,  9,  12,  18,  24. 

26.  What  is  the  quotient  of  f°f^|  divided  b 


T8Tof  16  J  of 

Am. 


^f  87.  We  have  seen  (IT  69)  that  fractions  having  differ 
ent  denominators,  as  thirds,  sevenths,  elevenths,  &c.,  cnn- 
not  be  added  and  subtracted  until  they  are  changed  to  equal 
fractions,  having  a  common  denominator — a  process  which  is 
often  somewhat  tedious.  To  obviate  this  difficulty,  Decimal 
fractions  have  been  devised,  founded  on  the  Arabic  system  of 
notation. 


IT  87.  DECIMAL  FRACTIONS.  109 

If  a  unit  or  whole  thing  be  divided  into  10  equal  parts,  each 
of  those  parts  will  be  1  tenth,  thus,  T\y  of  1  =  TV  ;  and  if  each 
tenth  be  divided  into  10  equal  parts,  the  10  tenths  will  make 
100  parts,  and  each  part  will  be  1  hundredth  of  a  whole 
thing,  thus,  y1^  of  -^=^-3.  In  like  manner,  if  each  hun 
dredth  be  divided  into  ten  equal  parts,  the  parts  will  be  thou 
sandths,  T\y  of  Tthj-  =  To-W  ancl  so  on-  Such  are  called 
Decimal  Fractions,  from  the  Latin  decem,  meaning  ten. 

Common  fractions,  then,  are  the  common  divisions  of  a 
unit  or  whole  thing  into  halves,  thirds,  fourths,  or  any  num 
ber  of  parts  into  which  we  choose  to  divide  it. 

Decimal  fractions  are  the  divisions  of  a  unit  or  whole 
thing  first  into  10  equal  parts,  then  each  of  these  into  10 
other  equal  parts,  or  hundredths,  and  each  hundredth  into  10 
other  equal  parts,  or  thousandths,  and  so  on. 

The  parts  of  a  unit,  thereby,  increase  and  decrease  in  a  ten 
fold  proportion  in  the  same  mariner  as  whole  numbers. 

The  following  examples  will  shoiv  the  convenience  of  deci- 
mal  fractions. 

1.    Add  together  -f^  and  y1^. 

SOLUTION.  —  Since  1  tenth  makes  10  hundredths,  we  may  reduce 
the  tenths  to  hundredths  by  annexing  a  cipher  which,  in  effect,  multi 
plies  them  by  10. 

Thus,  -^-  =  20  hundredths, 
and  adding  15  hundredths, 


we  have  35  hundredths, 

2.    From  -f^  take  yVkV 

SOLUTION.  —  We  reduce  the  36  hun- 

36  __3(3Q  thousandths       dredths  to  thousandths  by  annexing  a 

187  thousandths!     cipher  to  multiply  it  by  10.    Then  sub 

tracting   and    borrowing   as   in    whole 

173  thousandths,     numbers,  we  have  left  173  thousandths, 


Questions.  —  ^[  87.  Wkat  are  fractions?  "What  occasions  the 
chief  difficulty  in  operations  with  common  fractions?  To  what  has  this 
difficulty  led  ?  What  are  decimal  fractions  ?  Why  are  they  so  called  ? 
What  are  the  divisions  and  subdivisions  of  a  unit  in  decimal  fractions? 
and  what  are  the  parts  of  the  1st,  2d,  &c.,  divisions  called?  With  what 
system  of  notation  do  these  divisions  of  a  unit  correspond  ?  What  is  the 
law  of  increase  and  decrease  in  the  Arabic  system  of  notation  ?  What, 
then,  do  you  say  of  the  increase  of  the  whole  numbers,  and  of  the  parts  ? 
How  do  these  divisions  of  a  unit  in  decimal  fractions  differ  from  the 
divisions  of  a  unit  in  common  fractions?  Give  examples  showing  the 
superiority  of  decimal  fractions. 
10 


110  DECIMAL   FRACTIONS.  H  88. 

NOTE.  —  The  pupil  will  notice,  that  in  thus  reducing  the  fraction 
$&,  the  -j-gg-  makes  60  thousandths,  and  the  •$&  makes  300  thou 
sandths. 

Notation  of  Decimal  Fractions. 

If  88.  1.  Let  it  be  required  to  find  the  amount  of  3S5-&  + 
Ifrj7^  -f-  ^TGTJTJ  4~  yo"  ffiy»  anc'  express  the  fractions  decimally. 

SOLUTION.  —  Since  1  hundred  integers  make  10  tens,  1  ten  10 
units,  1  unit  10  tenths,  1  tenth  10  hundredths,  &c.,  decreasing  uni- 
formli'  from  left  to  right,  we  may  write  down  the  numerators  of  the 
fractions,  placing  tenths  after  units,  hundredths  after  tenths,  and  so 
on,  in  this  way  indicating  their  values  without  expressing  their  denom 
inators.  We  place  a  point  (')  called  the  Decimal  point,  or  Separatrix, 
on  the  left  of  tenths  to  separate  the  fraction  from  units,  or  whole 
numbers. 

a,  jo  35  j»  As   10  in   each   right 

g              "§  ^  Jg              ~  ^3  hand  make  1  in  the  next 

m  j  <u  g  m  „•"«  p  left    hand    column,    the 

12  w  -2  -5  "§  :§  "3  »  .3  ^  la  §  adding  and  carrying  will 

§  §  S  j^  -§  jS  §  §  J>  j=! 


-        be  the  same  throughout 


3  2  5'  5  325  '5  00      as  *n  whole  numbers- 

1  6'7  6        or,        1  6<7  8  0          The  reducing  to  a  com- 
4<3  7  9  4'3  7  0      mon  denominator,  it  will 


<0  2  5  '0  2  5      be  seen'  is  s 

up  the  vacant  places  with 

346'684          346'684      ciphers,  which  are  ornit- 

ted  in  the  first  operation 

without  affecting  the  result,  each  figure  being  written  in  its 
proper  place. 

The  denominator  to  a  decimal  fraction,  although  not  ex 
pressed,  is  always  understood,  and  is  1  with  as  many  cipher** 
annexed  as  there  are  places  at  the  right  hand  of  the  point 
Thus,  '684  in  the  last  example,  is  a  decimal  of  3  places  ;  con 
sequently  1,  with  3  ciphers  annexed,  (1000,)  is  its  propej 
denominator.  Any  decimal  may  be  expressed  in  the  form  of 
a  common  fraction  by  writing  under  it  its  proper  denominator. 
Thus,  '684,  expressed  in  the  form  of  a  common  fraction,  is 


Questions.  —  If  88.  Have  decimal  fractions  numerators  and  de 
nominators,  and  both  expressed?  How  can  you  write  the  numerators 
so  as  to  indicate  the  value  of  the  fraction,  without  expressing  the  denom- 
inpvtor?  How  can  the  denominator  be  known,  if  it  is  not  expressed? 
What  is  the  separatrix  and  its  use?  How  can  a  decimal  be  expressed 
in  form  of  a  common  fraction  ?  How  many,  and  what  advantages  have 
decimal  over  common  fractions  ? 


DECIMAL   FRACTIONS. 


Ill 


NOTE. — Tie  decimal  point  can  never  be  safely  omitted  in  opera 
tions  with  decimals.  Hence, 

Decimal  fractions  have  two  advantages  over  common  frac 
tions. 

First, — They  are  more  readily  reduced  to  a  common 
denominator. 

Second,  —  They  may  be  added  and  subtracted  like  whole 
numbers,  without  the  formal  process  of  reducing  them  to  a 
common  denominator. 

The  names  of  the  places  to  ten-millionths,  and,  generally 
how  to  read  or  write  decimal  fractions,  may  be  seen  from  the 
following 

TABLE. 


<! 

t-r 
)<=(      Cff^ 


H 


P  ^  f  3d  place. 
|."g*J    2d  place. 

3  CD"  1  1st  place. 

B§                       ' 

q-  q    o|    <=| 

-II  II  II  II 

CO  1C 
t^  Ol  <l 

||   ||   ||  Hundreds. 
Tens. 
Units. 

y  f  1st  place. 

o  c:  o  oo  os  o  o  o>  Tenths. 

g       2d  place. 

o  co  o  oi  GO  01  en     Hundredths. 

jg"      3d  place. 

o      o  o  on  o         Thousandths. 

£L  -  4th  place. 

O        O  O  CO 

Ten-Thousandths. 

^"d     5th  place. 

o      o  ro 

Hundred-Thousandths 

£     6th  place. 

O        CO 

Millionths. 

*   (  7th  place. 

o* 

Ten-Millionths 

O5  IO  <I  00  Oi 
ig^P    §g? 

•  s  p  oco 

:    :    P- 

III 

P    P-  tO 

^g  s 

K  CD 

?  s-s^ 

O^  t>  c    S 

S     CD     =« 

**~   ^—  J  ^       1 

P     P- 

WF^S 

B-f 

S    0    CD     0 

?r  M 

5  0  p*  c 

CO 

CD       CO       j     ^-    h^ 

Qj   •         H^      ^ 

?T       o    P- 

rP   C- 
P   ^ 

P- 

r 

112  DECIMAL  FRACTIONS.  H"  89,  90 

IT  89.  To  read  Decimals.  —  As  every  fraction  has  a  nu 
merator  and  a  denominator,  to  read  the  decimal  fraction,  of 
which  the  denominator  is  not  expressed,  requires  two  enumer 
ations,  —  one  from  left  to  right  to  ascertain  the  denominator, 
that  is,  the  name  or  denomination  of  the  parts,  and  another 
from  right  to  left,  to  ascertain  the  numerator,  that  is,  the  num 
ber  of  parts. 

Take,  for  instance,  the  fraction  '00387.  We  begin  at  the 
first  place  on  the  right  hand  of  the  decimal  point  and  say,  as 
in  the  table,  tenths,  hundredths,  &c.,  to  the  last  figure,  which 
we  ascertain  to  be  hundred-thousandths,  and  that  is  the  name 
or  denomination  of  the  fraction.  Then,  to  know  how  many 
there  are  of  this  denomination,  that  is,  to  determine  the  nu 
merator,  we  begin  as  in  whole  numbers,  and  say  units,  (7,) 
tens,  (8,)  hundreds,  (3,)  which  being  the  highest  significant 
figure,  we  proceed  no  further  ;  and  find  that  we  have  387 
hundred-thousandths,  (TTT3uVin7')  tne  numerator  being  387. 

In  this  way  a  mixed  number  may  be  read  as  a  fraction. 
Take  25'634.  Beginning  at  the  first  place  at  the  right  of  the 
point  we  have  tenths,  hundredths,  thousandths,  (the  lowest 
denomination  ;)  then  beginning  at  the  right  with  4,  we  say, 
units,  tens,  &c.,  as  in  whole  numbers,  and  find  that  we  have 
25634  thousandths,  which,  expressed  as  a  common  fraction,  is 


5F  !MK     To  ivrite  Decimal  Fractions. 

I.  Write  the  given  decimal  in  such  a  manner  that  each 
figure  contained  in  it  may  occupy  the  place  corresponding  to 
its  value. 

II.  Fill  the  vacant  places,  if  any,  with  ciphers,  and  put 
the  decimal  point  in  its  proper  place. 

Forty-six  and  seven  tenths  =  46-^  =  46'7. 

Write  the  following  numbers  in  the  same  manner  : 
Eighteen  and  thirty-four  hundredths. 
Fifty-two  and  six  hundredths. 

Nineteen  and  four  hundred  eighty-seven  thousandths. 
Twenty  and  forty-two  thousandths. 
One  and  five  thousandths. 
135  and  3784  ten-thousandths. 
9000  and  342  ten-thousandths. 


Questions.  —  Tf  89.  To  read  decimals  requires  what  ?  how  made 
and  for  what  purposes  ?  How  may  a  mixed  number  be  read  as  a  frau 
tion? 

If  9O      What  is  the  rule  for  writing  decimal  fractions  ? 


IT  91.  DECIMAL  FRACTIONS.  113 

10000  and  15  ten-thousandths. 

974  and  102  millionths. 

320  and  3  tenths,  4'hundredths,  and  2  thousandths. 

500  and  5  hundred-thousandths. 

47  millionths. 

Four  hundred  and  twenty-three  thousandths. 


Reduction  of  Decimal  Fractions. 

1TO1.  The  value  of  every  figure  is  determined  by  its  place 
from  units.  Consequently,  ciphers  annexed  to  decimals  do  not 
alter  their  value,  since  every  significant  figure  continues  to 
possess  the  same  place  from  unity.  Thus,  '5,  '50,  '500,  are 
all  of  the  same  value,  each  being  equal  to  y5^,  or  J. 

But  every  cipher  prefixed  to  a  decimal  diminishes  it  tenfold, 
by  removing  the  significant  figures  one  place  further  from 
unity,  and  consequently  making  each  part  only  one  tenth  as 
large.  Thus,  '5,  '05,  '005,  are  of  different  value,  '5  being 
equal  to  -ffr,  or  J  ;  '05  being  equal  to  yf T,  or  ^ ;  and  '005 
being  equal  to  f^,  or  ^. 

A  whole  number  is  reduced  to  a  decimal  by  annexing  ci 
phers  ;  to  tenths  by  annexing  1  cipher,  since  this  is  multiply 
ing  by  10 ;  to  hundredths  by  annexing  two  ciphers,  &c. 
Thus,  if  1  cipher  be  annexed  to  25  it  will  be  25'0,  (250 
tenths  ;)  if  2  ciphers,  it  will  be  25'00,  (2500  hundredths.) 

Several  numbers  may  be  reduced  to  decimals,  having  the 
same  or  a  common  denominator,  by  annexing  ciphers  till  all 
have  the  same  number  of  decimal  places.     Thus,  15'7,  '75, 
12^183,  9'0236  and  17'  are  reduced  to  ten  thousandths,  the 
lowest  denominator  contained  in  them,  as  follows : 
15'7        =  15'7000,  annexing  three  ciphers. 
'75      =      '7500,         "         two 


Questions.  —  1[  91.  How  is  the  value  of  every  decimal  figure  de 
termined  ?  How  do  ciphers  at  the  left  of  a  decimal  affect  its  value  ?  at 
the  right,  how?  In  the  fraction  '02643,  what  is  the  value  of  the  4?  of 
the  2  ?  How  does  the  0  affect  the  value  of  the  fraction  ?  In  the  fraction 
'15012  is  the  value  of  each  significant  figure  affected  by  the  0?  if  not, 
point  out  the  difference,  and  wherefore  ?  If  from  the  fraction  '8634  we 
withdraw  the  6,  leaving  the  fraction  to  consist  of  the  other  three  figures 
only,  how  much  should  we  deduct  from  its  value  ?  Demonstrate  by 
some  process  that  you  are  right.  How  may  integers  be  reduced  to  deci 
mals  ?  Reduce  46  to  thousandths ;  how  many  thousandths  does  the 
number  make  ?  How  may  several  numbers  be  reduced  to  decimals 
having  a  common  denominator  ?  To  what  denominator  should  they  all 
be  reduced? 

10* 


114  DECIMAL  FRACTIONS.  f  92. 

12' 183    =  124830,  annexing  one  cipher. 
9'0236  =    9'0236,  already  ten-thousandths. 
17'          =  17'0000,  annexing  four  ciphers. 

NOTE.  —  All  the  numbers  should  be  reduced  to  the  denominator  of 
the  one  having  the  greatest  number  of  decimal  places. 

EXAMPLES. 

1.  Reduce  7'25,  14<082,  2'3,  '00083,  and  25  to  a  common 
denominator. 

2.  Reduce  24,  3'02,  '425,  32'98762,  and  '3000001,  to  a 
common  denominator. 

IT  99.     To  change  or  reduce  Common  to  Decimal  Fractions. 

1.  A  man  has  |  of  a  barrel  of  flour ;  what  is  that,  expressed 
in  decimal  parts  ? 

As  many  times  as  the  denominator  of  a  fraction  is  contained 
in  the  numerator,  so  many  whole  ones  are  contained  in  the 
fraction.  We  can  obtain  no  whole  ones  in  f,  because  the 
denominator  is  not  contained  in  the  numerator.  We  may, 
however,  reduce  the  numerator  to  tenths,  by  annexing  a  cipher 
to  it,  (which,  in  effect,  is  multiplying  it  by  10,)  making  40 
tenths,  or  4;0.  Then,  as  many  times  as  the  denominator,  5, 
jj  contained  in  40,  so  many  tenths  are  contained  in  the  frac 
tion.  5  into  40  goes  8  times,  and  no  remainder. 

Ans.  '8  of  a  bushel. 

2.  Express  |  of  a  dollar  in  decimal  parts. 

OPERATION.  The  numerator,  3,  reduced  to 

Num.  tenths,  is  f  $,  3'0,  which,  divided 

Denom.  4 )  3<0  ( '75  of  a  dollar.     b7  the  denominator  4,  the  quo- 
go  tient  is  7  tenths,  and  a  remainder 
of  2.     This  remainder  must  now 
be  reduced  to  hundredlhs  by  an 
nexing  another  cipher,  making  20 

20  hundredths.       Then,    as     many 

times   as   the   denominator,  4,  is 

contained  in  20,  so  many  hundredths  also  may  be  obtained.  4  into  20 
5  times,  and  no  remainder.  |  of  a  dollar,  therefore,  reduced  to  deci 
mal  parts  is  7  tenths  and  5  hundredths  ;  that  is,  '75  of  a  dollar. 

Questions.  —  U"  92.  To  what  is  the  value  of  every  fraction  equal ? 
(If  64.)  How  is  a  common  fraction  reduced  to  a  decimal  ?  Of  how 
many  places  must  the  quotient  consist?  When  there  are  not  so  many 
places  how  is  the  deficiency  to  be  supplied?  Repeat  the  rule.  What  is 
the  course  of  reasoning  advanced  to  establish  this  rule  ? 


IF  92.  DECIMAL  FRACTIONS.  115 

3.    Reduce  ^g  to  a  decimal  fraction. 

The  numerator  must  be  reduced  to  hundredth*,  by  annexing 
two  ciphers,  before  the  division  can  begin. 

66 )  4*00  ( '0606  4-,  the  Answer. 

396 

As  there  can  be  no  tenths,  a 

cipher  must  be  placed  in  the  quo 
tient,  in  tenths'  place. 


NOTE.  —  -fe  cannot  be  reduced  exactly;  for,  however  long  the 
division  be  continued,  there  will  still  be  a  remainder.*  It  is  sufficiently 
exact  for  most  purposes,  if  the  decimal  be  extended  to  three  or  four 
places. 


*  Decimal  figures,  which  continually  repeat,  like  '06,  in  this  example,  are 
called  Repetends,  or  Circulating-  Decimals.  If  only  one  figure  repeats,  as 
'3333  or  '7777,  &c.,  it  is  called  a  single  repetend.  If  two  or  morejigures  cir 
culate  alternately,  as  '060606,  '234234234,  &c.,  it  is  called  a  compound  repetend. 
If  other  figures  arise  before  those  which  circulate,  as  '743333,  '143010101,  &c., 
the  decimal  is  called  a  -mixed  repetend. 

A  single  repetend  is  denoted  by  writing  only  the  circulating  Jigure  with  a 
point  over  it :  thus,  '3,  signifies  that  the  3  is  to  be  continually  repeated,  forming 
an  infinite  or  never-ending  series  of  3s. 

A  compound  repetend  is  denoted  by  a  point  over  the  first  and  last  repeating 
Jigures:  thus,  '234  signifies  that  234  is  to  be  continually  repeated. 

It  may  not  be  amiss,  here,  to  show  how  the  value  of  any  repetend  may  be 
found,  or,  in  other  words,  how  it  may  be  reduced  to  its  equivalent  vulgar  frac 
tion. 

If  we  attempt  to  reduce  ^  to  a  decimal,  we  obtain  a  continual  repetition  of 
the  figure  1 :  thus,  '11111,  that  is,  the  repetend  'i.  The  value  of  the  repetend 
ci,  then,  is  £  ;  the  value  of  '222,  &c.,  the  repetend  '2,  will  evidently  be  twice  as 
much,  that  is,  §-.  In  the  same  manner,  3  =^,  and  '4  =£,  and  '5  =  £,  and  so 
on  to  9,  which  =  -|-=  1. 

1.  What  is  the  value  of  '8  ?  Ans.  f-. 

2.  What  is  the  value  of  '6  1     Ans.  f  =  f .     What  is  the  value  of  <3  ?  -— 
of '7?  — —  of<4?  of '5?  of '9?  of'i? 

If  ^  be  reduced  to  a  decimal,  it  produces  '010101 ,  or  the  repetend  (6i .  The 
repetend  '02,  being  2  times  as  much,  must  be  •£§  and  '03=  ^-,  and '48,  being 
48  times  as  much,  must  be  |-f-,  and  '74  =-g-f-,  &c. 

If  -§^ff  be  reduced  to  a  decimal,  it  produces  '00  i  ;  consequently,  '002  = 
g-^-g-,  and  '037  =  ^5-,  and  '425  =  ij-f  %,  &c.  As  this  principle  will  apply  to 
any  number  of  places,  we  have  this  general  RULE  for  reducing  a  circulating 
decimal  to  a  common  fraction. 

Make  the  given  repetend  the  numerator,  and  the  denominator  will  be  as 
many  9s  as  there  are  repeating- Jigures. 


116  DECIMAL  FRACTIONS.  f  93. 

From  tha  foregoing  examples  we  may  deduce  the  following 
general 

RULE. 

To  reduce  a  common  to  a  decimal  fraction.  —  Annex  one  or 
more  ciphers,  as  may  be  necessary,  to  the  numerator,  and 
divide  it  by  the  denominator.  If  then  there  be  a  remainder, 
annex  another  cipher,  and  divide  as  before,  and  so  continue  to 
do,  so  long  a's  there  shall  continue  to  be  a  remainder,  or  until 
the  fraction  shall  be.  reduced  to  any  necessary  degree  of  exact 
ness. 

The  quotient  will  be  the  decimal  required,  which  must  con 
sist  of  as  many  decimal  places  as  there  are  ciphers  annexed 
to  the  numerator  ;  and,  if  there  are  not  so  many  figures  in  the 
quotient,  the  deficiency  must  be  supplied  by  prefixing  ciphers. 

EXAMPLES    FOR    PRACTICE. 


4.  Reduce  J,  J,  ¥^y,  and  y^g-  to  decimals. 

Ans.  '5;  '25;  '025;  '00797  +. 

5.  Reduce  f£,  -^^  -rfasi  and  ^VVF  to  decimals. 

Ans.  '692  +  ;  '003;  '0028  +  ;  '000183  +. 

6.  Reduce  jjf  ,  ^V,  ^  to  decimals. 

7.  Reduce  £,  ¥5¥,  ^f-F,  J,  f,  -Jy,  -fa,  -^-^  to  decimals. 

8.  Reduce  f,  f  ,  f  ,  £,£,§,  f  ,  &,  ^,  &  to  decimals. 


Federal  Money. 

IT  93.     Federal  Money  is  the  currency  of  the  United  States 

The  unit  of  English  money  is  the  pound  sterling,  which  is 
divided  into  20  equal  parts,  (twentieths,)  called  shillings ; 

3.  What  is  the  vulgar  fraction  equivalent  to  '704  1  Ans.  ^^. 

4.  What  is  the  value  of  '003?  {014?  '324?  [0102i  ?  

'2463  ?   '002103  ?  Ans.  to  last,  jn?j&W 

5.  What  is  the  value  of  '43  ? 

In  this  fraction,  the  repetend  begins  in  the  second  place,  or  place  of  hun- 
dredths.  The  first  figure,  4,  is  y^,  and  the  repetend,  3,  is  |-  of  T^.,  that  is, 
T^J-  ;  these  two  parts  must  be  added  together.  iV~f~  !r&==%ij=:=^l5)  Ans. 
Hence,  to  find  the  value  of  a  mixed  repetend,  —  Find  the  value  of  the  two 
parts,  separately,  and  add  them  together. 

6.  What  is  the  value  of '153?  ^/>v  +  -<&v==U%  =  ilfc,  A™' 

7.  What  is  the  value  of  '0047  ?  Ans.  ^f  TT- 

8.  What  is  the  va.  je  of  '138  ?  (16  ?  '4123  ? 

It  is  plain,  that  circulates  may  be  added,  subtracted,  multiplied,  and  divided, 
by  first  reducing  them  to  their  equivalent  vulgar  fractions. 


IT  93.  DECIMAL  FRACTIONS.  117 

each  shilling  is  divided  into  12  parts  called  pence  ;  a  penny 
being  S^TT  °f  a  pound.  Each  penny  again  is  divided  into  4 
parts  called  farthings,  a  farthing  being  ^^  of  a  pound.  These 
divisions,  therefore,  are  like  those  of  common  fractions,  and 
the  same  difficulties  occur  in  operations  with  English  money 
as  with  common  fractions. 

The  unit  of  Federal  money  is  the  Dollar,  divided  into  10 
parts  called  dimes,  from  a  French  word  meaning  tenth  (of  a 
dollar) ;  each  dime  into  10  parts  called  cents,  from  the  French 
for  hundredth  (of  a  dollar) ;  and  each  cent  into  10  parts  called 
mills,  from  the  French  for  thousandth  (of  a  dollar).  These 
divisions  of  the  money  unit  are  like  those  of  decimal  fractions. 
Our  money,  then,  has  this  advantage  over  the  English,  viz., 
that  operations  in  it  are  as  in  whole  numbers,  and  we  shall 
therefore  consider  it  in  connection  with  Decimals. 

The  denominations  of  Federal  money  are  eagles,  dollars, 
dimes,  cents,  and  mills. 

TABLE. 

10  mills  make  1  cent. 

10  cents  (  =  100  mills)  1  dime. 

10  dimes  ( =  100  cents  =  1000  mills)      1  dollar. 

10  dollars  1  eagle. 

NOTE. — Coin  is  a  piece  of  metal  stamped  with  certain  impressions 
to  give  it  a  legal  value,  and  also  to  serve  as  a  guarantee  for  its  weight 
and  purity. 

The  mill  is  so  small  that  it  is  not  usually  regarded  in  busi 
ness.  The  eagle  is  merely  the  name  of  a  gold  coin  worth  10 
dollars.  Dimes  are  read  as  10s  of  cents.  Federal  money, 
then,  is  calculated  in  dollars  and  cents,  and  accounts  are  kept 
in  these  denominations. 

A  character,  $,  which  may  be  regarded  a  contraction  of  U. 
S.,  placed  before  a  number,  signifies  that  it  is  Federal,  or 
U.  S.  money. 

Questions.  —  IT 93.  "What  is  Federal  money?  What  is  the  unit 
of  English  money?  What  are  its  denominations?  and  what  are  they 
like  ?  What  is  the  unit  of  Federal  money  ?  how  divided  ?  and  whence 
the  names  of  the  divisions  ?  What  advantages  has  Federal  over  English 
money  ?  Repeat  the  table.  What  is  the  eagle  ?  How  are  dimes  read  ? 
In  what  then  is  Federal  money  calculated,  and  accounts  kept  ?  What 
is  coin?  What  is  the  character  for  U.  S.  money,  and  where  placed? 
Where  is  the  decimal  point  placed?  Where  and  how  many  are  the 
places  for  cents  ?  for  mills  ?  Why  more  places  for  cents  than  for  mills  ? 
If  the  sum  be  but  8  cents  how  may  it  be  written  ?  if  three  mills  only, 
how  ?  How  are  5  mills  usually  written  ? 


118  DECIMAL  FRACTIONS.  IT  94 

As  the  dollar  is  the  unit  of  Federal  money,  the  decimal 
point  is  placed  at  the  right  hand  of  dollars  ;  arid  since  dimef 
(tenths)  and  cents  (hundredths)  are  read  together  as  cents 
the  first  two  places  at  the  right  hand  of  the  point  expres? 
cents,  and  the  third,  mills,  (thousandths.)  Thus,  25  dollars 
78  cents,  and  6  mills,  are  written,  $25'786. 

If  there  be  no  dimes,  (tenths,  or  10s  of  cents,)  that  is,  if 
the  cents  are  less  than  10,  a  cipher  is  put  in  the  place  of 
tenths ;  thus,  8  cents  are  written  $'08 ;  and  if  there  are  only 
mills,  ciphers  must  be  put  in  the  place  of  tenths  and  hun 
dredths  ;  thus,  5  mills  are  written  $'005.  But  5  mills  are 
usually  expressed  as  half  a  cent ;  thus,  12  cents  5  mills,  are 
written  12|  cents,  or  $'12J. 


Reduction  of  Federal  Money. 

If  94.  It  is  evident  that  dollars  are  reduced  to  cents  in  the 
same  manner  as  whole  numbers  are  reduced  to  hundredths, 
by  annexing  two  ciphers  ; 

To  mills  or  thousandths,  by  annexing  three  ciphers. 

On  the  contrary, 

Mills  are  reduced  to  cents  by  cutting  off  the  right  hand 
figure ; 

To  dollars,  by  cutting  off  three  figures  from  the  right, 
which  is  dividing  by  1000,  (H  41.) 

Cents  are  reduced  to  dollars  by  cutting  off  two  figures  from 
the  right,  which  is  dividing  by  100. 

EXAMPLES. 

1.    Reduce  $34  to  cents.  2.    Reduce  48143  mills  to 

Ans.  3400  cents.  dollars.          Ans.  $48' 143. 
3.   Reduce  $40'06|  to  mills.         4.    Reduce  48742  cents  to 

Ans.  40065  mills.  dollars.          Ans.  $487'42. 
5.    Reduce  $16  to  mills.  6.    Reduce    125    mills    to 

Ans.  16000  mills.  cents.  $'12J. 

7.    Reduce  $'75  to  mills.  8.    Reduce  2064J  cents  to 

Ans.  750  mills.  dollars.          Ans.  $20'64|. 
9.    Reduce  $'007  to  mills.          10.    Reduce  9  cents  to  dol- 

Ans.  7  mills.  lars.  Ans.  $'09. 

Questions.  —  1f  94.  How  are  dollars  reduced  to  cents?  to  mills? 
cents  to  mills  ?  mills  to  cents  ?  to  dollars  ?  cents  to  dollars  ? 


IT  95.  DECIMAL  FRACTIONS.  119 

Addition  and  Subtraction  of  Decimal  Frac 
tions. 

^f  95.  As  the  value  of  the  decimal  parts  of  a  unit  vary 
in  a  tenfold  proportion  like  whole  numbers,  the  addition  and 
subtraction  of  decimal  fractions,  and  of  Federal  money,  may  be 
performed  as  in  whole  numbers. 

1.    What   is    the    amount         2.    From  765<06  take  27- 
of    14'68,   9'045,    38'5,   and    '6895. 
9<0025  ? 

SOLUTION.  —  As  numbers  of  the  same  denomination  only  can  be 
added  together,  (*J[  12,)  or  subtracted  from  each  other,  the  several 
numbers  in  each  of  these  examples  must  be  reduced  to  the  lowest 
denomination  contained  in  any  one  of  the  numbers,  (*fl  91,)  which  is 
ten-thousandths,  when  the  operation  may  be  performed  as  in  simple 
numbers. 
FIRST  OPERATION.  Or  if  only  like  denominations  FIRST  OPERATION. 

1  1'6SOO  are  written  under  each  other,     765'0600 

9*0450  these  alone  will  be  added  to-       27'6895 

oo<  "nf\f\  gether,   or   subtracted    from 

* 


°ther'   and    the    °Pera'     737'3705'    AnS' 
___  tions  may  be  performed  with- 

71'2275    Ans       out  tne  f°rmalitv  of  reducing  to  a  common  denom 
inator,  since  .the  ciphers,  by  which  the  reduction 
is  effected,  make  no  difference  in  the  result  :  thus, 

SEOOND    OPERATION.  NOTE.  -  As  the    deci-      SECOND    OPERATION 

14'68  mal  point  is  at  the  right  765'06 

9'045  °f  units?  which  are  writ-  27'6895 

ten  under  each  other,  the 


9'0025  P0*n    n    i        u  "  737'3705 

_  rectly  below  the  points  in 

71  '2275  the  several  numbers.     Hence, 

To  add  or  subtract  decimal  fractions, 

RULE. 

Write  the  numbers  under  each  other,  tenths  under  tenths, 
hundredths  under  hundredths,  &c.,  according  to  the  value  of 
their  places  ;  add  or  subtract  as  in  simple  numbers,  and  point 
off  in  the  result  as  many  places  for  decimals  as  are  equal  to 

Questions.  —  IF  95.  Why  can  addition  and  subtraction  of  deci 
mals  be  performed  as  in  whole  numbers  ?  What  numbers  only  can  be 
added  and  subtracted  ?  How  is  this  effected  by  the  first  operations  ? 
How  by  the  second  ?  How  do  you  write  down  decimals  for  addition  ? 
How  for  subtraction  ?  Where  place  the  point  in  the  results  ?  Repeat 
the  rule  How  do  you  prove  addition  of  decimals  ?  How  subtraction? 


120  DECIMAL  FRACTIONS.  IT  95. 

the  greatest  number  of  decimal  places  in  any  of  the  given 
numbers. 

PROOF.  —  The  same  as  in  the  addition  and  subtraction  of  simple 
numbers 

EXAMPLES    FOR    PRACTICE. 

3.  A  man  sold  wheat  at  several  times  as  follows,  viz., 
13'25  bushels,  8'4  bushels,  23'051  bushels,  6  bushels,  and  '75 
of  a  bushel ;  how  much  did  he  sell  in  the  whole  ? 

Am.  51'451  bushels. 

4.  What  is  the  amount  of  429,  21T3^,  355^^,  1T^,  and 
Ift  ?  Am.  808^ .  or  808'  143. 

5.  What  is  the  amount  of  2  tenths,  80  hundredths,  89  thou 
sandths,  6  thousandths, 9  tenths,  and  5  thousandths? 

Ans.  2. 

6.  What  is  the  amount  of  three  hundred  twenty -nine  and 
seven  tenths,  thirty-seven  and  one  hundred  sixty-two  thou 
sandths,  and  sixteen  hundredths  ? 

7.  From  thirty-five  thousand  take  thirty-five  thousandths. 

Ans.  34999<965. 

8.  From  5'83  take  4'2793.  Ans.  1'5507. 

9.  From  480  take  245'0075.  Ans.  2349925. 

10.  What  is  the  difference  between  1793' 13  and  817*05 
693?  Ans.  976'07307. 

11.  From  4^  take  2^.        Remainder,  1£&,  or  1<98. 

12.  What  is  the  amount  of  29ft,  374linr$Tnnr,  97-flfcft, 
315^*  27,  and  100ft  ?  Ans'  942^957009. 

Examples  in  Federal  Money  can  evidently  be  performed  in  the 
same  way. 

1.  Bought  1  barrel  of  flour  for  6  dollars  75  cents,  10  pounds 
of  coffee  for  2  dollars  30  cents,  7  pounds  of  sugar  for  92  cents, 
1  pound  of  raisins  for  12J  cents,  and  2  oranges  for  6  cents  • 
what  was  the  whole  amount?  Ans.  $10' 155. 

2.  A  man  is  indebted  to  A,  $237<62;  to  B,  $350;  to  C, 
$86' 12i  ;  to  D,  $9'62i  ;  and  to  E,  $0<834;  what  is  the  amount 
of  his  debts?  jte.8684'204. 

3.  A  man  has  three  notes  specifying  the  following  §ums, 
viz.,  three  hundred  dollars,  fifty  dollars  sixty  cents,  and  nine 
dollars  eight  cents ;  what  is  the  amount  of  the  three  notes  ? 

4.  What  is  the  amount  of  $56' 18,  $7'37J,  $280,  $0'287 
$17,  and  $90'413  ?  Ans.  $451'255. 


H96.  DECIMAL  FRACTIONS.  121 

5.  Bought  a  pair  of  oxen  for  $76'50,  a  horse  for  $85,  and 
a  cow  for  $17 '25;  what  was  the  whole  amount? 

A?is.  $178'75. 

6.  Bought  a  gallon  of  molasses  for  28  cents,  a  quarter  of 
tea  for  37 1  cants,  a  pound  of  saltpetre  for  24  cents,  2  yards 
of  broadcloth  for  11  dollars,  7  yards  of  flannel  for  1  dollar  62  J 
cents,  a  skein  of  silk  for  6  cents,  and  a  ^tick  of  twist  for  4 
cents  ;  how  much  for  the  whole  ?  Ans.  $13'62. 

7.  A  man  bought  a  cow  for  eighteen  dollars,  and  sold  her 
again  for  twenty-one  dollars  thirty-seven  and  a  half  cents  ; 
how  much  did  he  gain  ?  Ans.  $3'375. 

8.  A  man  bought  a  horse   for  82  dollars,  and  sold  him 
again  for  seventy-nine  dollars  seventy-five  cents ;  did  he  gain 
or  lose  ?  and  how  much  ?  Ans.  He  lost  $2'25. 

9.  A  merchant  bought  a  piece  .of  cloth  for   $176,  which 
proving  to  have  been  damaged,  he  is  willing  to  lose  on  it 
$16'50  ;  what  must  he  have  for  it?  Ans.  $159'50. 

10.  A  man  sold  a  farm  for  $5400,  which  was  $725'37J 
more  than  he  gave  for  it ;  what  did  he  give  for  the  farm  ? 

11.  A  man,  having  $500  dollars,  lost  S3  cents;  how  much 
had  he  left?  Ans.  $499' 17. 

12.  A  man's  income  is   $1200  a  year,  and   he   spends 
$800'35 ;  how  much  does  he  lay  up  ? 

13.  Subtract  half  a  cent  from  seven  dollars. 

Bern.  $6<99|. 

14.  How  much  must  you  add  to  $16'82  to  make  $25? 

15.  How  much  must  you  subtract  from   $250,  to  leave 
$87' 14? 

16.  A  man  bought  a  barrel  of  flour  for  $6'25,  7  pounds  of 
coffee  for  $1'41,  he  paid  a  ten  dollar  bill ;  how  much  must  he 
receive  back  in  change  ?  Ans.  $2'34. 


Multiplication  of  Decimal  Fractions. 

IF  96.     1.    Multiply '7  by '3. 

'7  =  A  and  '3  =  T3<i,  then  ^XT3T==^V  =  '21,  Ans. 
We  here  see  that  tenths  multiplied  into  tenths  produce  hun- 
dredths,  just  as  tens,  (70,)  into  tens,  (30,)  make  hundreds, 
(2100.)  We  may  write  down  the  numerators  decimally, 
thus: 

11 


122 


DECIMAL  FRACTIONS. 


IT  96. 


OPERATION. 
'7 


'21,  Am. 


The  21  must  be  hundredths  as  before.  The  number 
of  figures  in  the  product,  it  will  be  seen,  is  equal  to  the 
number  in  the  multiplicand  and  multiplier  ;  hence,  we 
have  as  many  places  for  decimals  in  the  product  as  there 
are  in  both  the  factors. 


NOTE.  —  The  correctness  of  the 
above  rulfe  may  be  illustrated  by  the 
annexed  diagram.  The  length  of  the 
plot  of  ground  which  it  represents 
may  be  regarded  10  feet  and  thf 
breadth  10  feet,  each  division  of  a 
line,  consequently,  being  one  tenth 
of  the  whole  line.  Multiplying  10 
by  10,  we  have  100  square  feet  in 
the  plot,  each  of  the  small  squares 
being  1  square  foot,  or  one  hun 
dredth  of  the  whole  plot.  Now  take 
the  part  encircled  by  the  black  lines, 
7  feet  ('7  of  the  whole  line)  long, 
and  3  feet  ('3  of  the  whole  line)  wide.  The  contents  are  81  square 
feet,  or  '21  of  the  whole  plot;  hence  the  product  of  '7  into  '3  is  '21 
as  above. 

2.    Multiply  '125  by  '03. 

Here,  as  the  number  of  significant  figures  in  the 

OPERATION.        product  is  not  equal  to  the  number  of  decimals  in 

<12o  both  factors,  the  deficiency  must  be  supplied  by  pre- 

403  fixing  ciphers,  that  is,  placing  them  at  the  left  hand. 

-  The  correctness  of  the  rule  may  appear  from  the 

'00375  Prod,     following  process  :   '125  is  TV2u5<y>  and  '03  is  ^  : 

now,    TVVn  X  Ttnr  =  TtfinfW  =  '00375,   the 

same  as  before. 

Hence,  To  multiply  decimal  fractions  , 

RULE. 

Multiply  as  in  whole  numbers,  and  from  the  right  hand  of 
the  product  point  off  as  many  figures  for  decimals  as  there 
are  decimal  places  in  the  multiplicand  and  multiplier  counted 
together,  and  if  there  are  not  so  many  figures  in  the  product, 
supply  the  deficiency  by  prefixing  ciphers. 

Questions.  —  ^f  96.  Tenths  X  tenths  produce  what  ?  Illustrate 
this  by  a  diagram.  To  what  must  the  number  of  places  in  the  product 
be  equal  ?  When  the  number  of  decimals  in  the  product  is  less  than  the 
number  in  both  factors  what  do  you  do  ?  How  can  you  tell  of  what 
name  or  denomination  will  be  the  product  of  one  given  decimal  mul 
tiplied  into  another  given  decimal,  without  going  through  the  process 
of  multiplication  ?  Of  what  denomination,  then,  will  be  the  product  of 
'46  X  '25  ?  of  '0005  X  '07  ? 


IT  97.  DECIMAL  FRACTIONS.  123 

EXAMPLES    FOR    PRACTICE. 

3.  Multiply  five  hundredths  by  seven  thousandths.  • 

Product,  '00035. 

4.  What  is  '3  of  116  ?  Ans.  34'8. 

5.  What  is  '85  of  3672  ?  Ans.  3121'2. 

6.  What  is  '37  fcf  '0563  ?  Ans.  '020831. 

7.  Multiply  572  by  '58.  Product,  331'76. 

8.  Multiply  eighty-six  by  four  hundredths. 

Product,  3'44. 

9.  Multiply  '0062  by  '0008. 

10.  Multiply  forty-seven  tenths  by  one  thousand  eighty-six 
hundredths.  Prod.  51'042. 

EXAMPLES    IN    FEDERAL    MONEY. 

IT  97.     1.    If  a  melon  be  worth  $'09,  what  is  '7  of  it 
worth  ?  (IF 77.)  Ans.  $'063. 

2.  What   will   250   bushels   of    rye   cost,   at   $'8SJ   per 
bushel  ? 

3.  What  is  the  value  of  87  barrels  of  flour,  at  $6'37-|  a 
barrel  ?  Ans.  $554'62|. 

4.  What  will  be  the  cost  of  a  hogshead  of  molasses,  con 
taining  63  gallons,  at  28J  cents  a  gallon  ?      Ans.  $17'955. 

5.  If  a  man  spend  12|  cents  a  day,  what  will  that  amount 
to  in  a  year  of  365  days?   What  will  it  amount  to  in  5  years? 

Ans.  It  will  amount  to  $228' 12|  in  5  years. 

6.  If  it  cost  $36'75  to  clothe  a  soldier  1  year,  how  much 
will  it  cost  to  clothe  an  army  of  17800  men  ? 

Ans.  $654150. 

7.  Multiply  $367  by  46. 

8.  Multiply  $0'273  by  8600.  Ans.  $2347'80. 

9.  At  $5'47  per  yard,  what  cost  8'3  yards  of  cloth.  ? 

Ans.  $45'401 

10.  At  $'07  per  pound,  what  cost  26'5  pounds  of  rice  ? 

Ans.  $1'855. 

11.  What  will  be  the  cost  of  thirteen  hundredths  of  a  ton 
of  hay,  at  $11  a  ton  ?  Ans.  $1'43. 

12.  What  will  be  the  cost  of  three  hundred  seventy-five 
thousandths  of  a  cord  of  wood,  at  $2  a  cord  ?  $'75. 

13.  If  a  man's  wages  be  seventy-five  hundredths  of  a  dol 
lar  a  day,  how  much  will  he  earn  in  4  weeks,   Sundays 
excepted  ?  Ans.  $18. 


124  DECIMAL  FRACTIONS.  H"  98. 

Division  of  Decimal  Fractions. 

IT  98.     1.    Divide  '21  by  '3.     «21  =  £fo,  and  '3  = -ft. 

Now  tffo-r-Tfr;  or  4f  of  -^  =  lBS  =  TV  It  appears, 
then,  that  hundredths  divided  by  tenths  give  tenths,  just  as 
hundreds  (2100)  divided  by  tens  (30)  giw  tens,  (70.)  The 
numerators  may  be  set  down  decimally,  and  the  division  per 
formed  as  follows :  — 

OPERATION.         The  7  must  be  tenths  as  before.      The   dividend, 
'3) '21          which  answers  to  the  product  in  multiplication,  con 
tains  two  decimal  places  ;  and  the  divisor  and  quotient, 
'7         which  answer  to  the  factors  in  multiplication,  ("[f  31,) 
together  contain  two  decimal  places.     Hence,  we  see 
that  the  number  of  decimal  places  in  the  quotient  is  equal  to  the  dif 
ference  between  the  number  in  the  dividend  and  divisor. 

2.  At  4'75  of  a  dollar  per  barrel,  how  many  barrels  of  flour 
can  be  bought  for  $31  ? 

OPERATION.  The  4'75  are  475  hundredths,  and,  since 

4'75)31'00(6'526-f-     the  dividend  and  divisor  must  be  of  the 
2850  same  denomination,  we  annex  2  ciphers  to 

31  and  it  becomes  3100  hundredths,  (Tf  91.) 
2500  Then  there  can  be  as  many  whole  barrels 

2375  bought  as  the  number  of  times  475  hun 

dredths  can  be  subtracted  from  3100  hun 
dredths.     The  6  barrels  thus   found   will 
950  cost.  2850  hundredths  of  a  dollar,  and  as 

250  hundreths  or  cents  remain,  it  will  buy 
Part  °^  anotner  Barrel,  which  we  find  by 
annexing  ciphers,  and  continuing  the  op- 
eration. 

We  now  see  that  there  are  5  decimal 
places  in  the  dividend,  counting  all  the  ciphers  that  are  annexed,  and  as 
there  are  but  two  in  the  divisor,  we  point  off  3  in  the  quotient.  There 
is  still  a  remainder  of  150,  which,  written  over  the  divisor,  (^[  36,) 
gives  ||-^  of  a  thousandth  of  a  barrel,  a  quantity  so  small  that  it 
may  be  neglected.  But  we  place  -f-  at  the  right  of  the  last  quotient 
figure,  to  show  that  there  is  more  flour  than  indicated  by  the  quotient. 

Ans.  6'526 -f  barrels. 

NOTE.  — It  is  sufficiently  exact  for  most  practical  purposes  to  carry 
the  division  to  three  decimal  places. 

3.  Divide  '00375  by '125. 

OPERATION.  The  divisor,  125,  in  375,  goes  3  times,  and 

125 )  '00375  ( '03     no  remainder.     We  have  only  to  place  the  deci- 

375  mal  point  in  the  quotient,  and  the  work  is  done. 

There  are  five  decimal  places  in  the  dividend ; 

000  consequently  there  must  be  five  in  the  divisor 


1T98.  DECIMAL  FRACTIONS.  125 

and  quotient  counted  together  ;  and  as  there  are  three  in  the  divisor, 
there  must  be  two  in  the  quotient  ;  and,  since  we  have  but  one  figure 
in  the  quotient,  the  deficiency  must  be  supplied  by  prefixing  a  cipher. 
The  operation  by  vulgar  fractions  will  bring  us  to  the  same  result. 
Thus,  '125  is  T\T2^,  and  '00375  is  y^W  :  now> 
=  '<>    the  same  M  before- 


4.  Divide  '75  by  '005. 

OPERATION.  SOLUTION.  —  We  cannot  divide  hundredths 

475  by  thousandths,  until  the  former  are  reduced  to 

JQ  thousandths  by  multiplying  by  10,  or  annexing 

—  _  one  cipher,  when  the  divisor  and  dividend  will 

'005  )  '750  be  of  the  same  denomination;  and  '005  is  con- 

--  tained  in  (can  be  subtracted  from)  '750,  150 

150,  Quot.     times,  the  quotient  being  a  whole  number. 

These  illustrations  will  establish  the  following 

RULE. 

I.  Reduce,  if  necessary,  the  dividend  to  the  lowest  denom 
ination  in  the  divisor,  divide  as  in  whole  numbers,  annexing 
ciphers  to  a  remainder  which  may  occur,  and  continuing  the 
operation 

II.  If  the  decimal  places  in  the  dividend  with  the  ciphers 
annexed  exceed  those  in  the  divisor,  point  off  the  excess  from 
the  right  of  the   quotient  as  decimals  ;  but  if  the  excess  is 
more  than  the  number  of  places  in  the  quotient,  supply  the 
deficiency  by  prefixing  ciphers. 

EXAMPLES    FOR    PRACTICE. 

5.  Divide  3156'293  by  25'  17.  Quot.  125'3-j-. 

6.  Divide  173948  by  '375.  Quot.  463861  +• 

NOTE.  —  The  pupil  will  point  off  the  decimal  places  in  the  quo 
tient  of  this  and  the  following  example,  as  directed  by  the  rule. 

7.  Divide  5737  by  13<3.  Quot.  431353. 

8.  What  is  the  quotient  of  2464'S  divided  by  '008? 

Ans.  308100. 


Questions.  —  ^  98.  Hundredths,  divided  by  tenths,  give  what? 
How  is  it  in  integers  ?  Exhibit  on  the  blackboard  the  process  of  divid 
ing  7  by  1  -25.  Why  do  you  annex  ciphers  to  the  7  ?  What  is  the  quo 
tient  ?  Why  pointed  thus'?  Give  a  demonstration  by  common  fractions, 
as  after  Ex.  3,  and  show  that  this  placing  of  the  point  is  right.  The 
sign  of  addition,  annexed  to  the  quotient,  is  an  indication  of  what  ? 
When  there  are  remainders,  to  how  many  places  should  the  division  be 
carried  ?  Why  not  to  more  places  ?  Repeat  the  rule  for  division 
11* 


126  DECIMAL  FRACTIONS.  1[  99 

9.  Divide  2  by  53'1.  Quot.  '037  -f . 

10.  Divide  '012  by  '005.  Quot.  2*4. 

11.  Divide  three  thousandths  by  four  hymdredths. 

Quot.  '075. 

12.  Divide  eighty-six  tenths  by  ninety-four  thousandths. 

13.  How  many  times  is  '17  contained  in  8? 

EXAMPLES    IN    FEDERAL    MONEY. 

f  99.  1.  Divide  $59<3S7  equally  among  8  men;  how 
much  will  each  man  receive  ? 

OPERATION. 

8)59'387 

Ans.  $7<423f ,  that  is,  7  dollars,  42  cents,  3  mills,  and  f  of 
another  mill.  The  g  is  the  remainder,  after  the  last  division, 
written  over  the  divisor,  and  expresses  such  fractional  part  of 
another  mill. 

For  most  purposes  of  business,  it  will  be  sufficiently  exact 
to  carry  the  quotient  only  to  mills,  as  the  parts  of  a  mill  are 
of  so  little  value  as  to  be  disregarded. 

2.  At  $'75  per  bushel,  how  many  bushels  of  rye  can  be 
bought  for  $141  ?  Ans.  188  bushels. 

3.  At  12 \  cents  per  lb.,  how  many  pounds  of  butter  may 
be  bought  for  $37  ?  Ans.  296  Ibs. 

4.  At  6J  cents  apiece,  how  many  oranges  may  be  bought 
for  $8  ?  Am.  128  orangesT 

5.  If  '6  of  a  barrel  of  flour  cost  $5,  what  is  that  per  bar 
rel?  /  Aiis.  88<333+. 

NOTE.  — If  the  sum  to  be  divided  contain  only  dollars,  or  dollan 
and  cents,  it  may  be  reduced  to  mills,  by  annexing  ciphers  before 
dividing ;  or,  we  may  first  divide,  annexing  ciphers  to  the  remainder, 
if  there  shall  be  any,  till  it  shall  be  reduced  to  mills,  and  the  result 
will  be  the  same. 

6.  If  I  pay  $468' 75  for  750  pounds  of  wool,  what  is  the 
value  of  1  pound?  Am.  $0'625;   or  thus,  $'62£. 

7.  If  a  piece  of  cloth,  measuring  125  yards,  cost  S181'25, 
what  is  that  a  yard  ?  Ans.  $1<45. 

8.  If  536  quintals  of  fish  cost  $1913'52,  how  much  is  "that 
a  quintal  ?  Ans.  $3'57 

9.  Bought  a  farm,  containing  84  acres,  for  $3213-;  what 
did  it  cost  me  per  acre  ?  Ans.  $3S'25. 

10.  At  $954  for  3816  yards  of  flannel,  what  is  that  a  yard  ? 

Ans.  $0'25. 


H100.  DECIMAL  FRACTIONS.  127 

11.    Bought  72  pounds  of  raisins  for  $8;  what  was  that  a 
pound?  Ans.  SO'lllJ;   or,  $0<lll-f. 

^  12.    Divide  812  into  200  equal  parts  ;  how  much  is  one  of 
the  parts  ?     ^  =  how  much  ?  Ans.  $'06. 

13.  Divide  $30  by  750.     ^  —  how  much  ? 

14.  Divide  $60  by  1200.     Tf ^  =  how  much  ? 

15.  Divide  $215  into  86  equal  parts  ;  how  much  will  one 
of  the  parts  be  ?     %^-  =  how  much  ? 


IT  1®0.    Review  of  Decimal  Fractions. 

Questions.  —  "What  are  decimal  fractions  ?  How  do  they  differ 
from  common  fractions  ?  How  can  the  proper  denominator  to  a  decimal 
fraction  be  known,  if  it  be  not  expressed  ?  What  advantages  have  deci 
mal  over  common  fractions  ?  How  is  the  value  of  every  figure  deter 
mined?  Describe  the  manner  of  numerating  and  reading  decimal  frac 
tions  ?  of  writing  them  ?  How  are  decimals,  having  different  denomi 
nators,  reduced  to  a  common  denominator?  How  may  any  whole 
number  be  reduced  to  decimal  parts?  How  can  any  mixed  number  be 
read  together,  and  the  whole  expressed  in  the  form  of  a  common  frac 
tion  ?  What  is  federal  money  ?  What  is  the  money  unit,  and  what  are 
its  divisions  and  subdivisions  ?  How  is  a  common  fraction  reduced  to  a 
decimal  ?  To  what  do  the  denominations  of  federal  money  correspond  ? 
What  is  the  rule  for  addition  and  subtraction  of  decimals  ?  —  multiplica 
tion  ?  —  division  ? 

EXERCISES. 

1.  A  merchant  had  several  remnants  of  cloth,  measuring  as 
follows,  viz. : 

7  |  yds.  ^       How  many  yards  in  the  whole,  and  what  would 
the  whole  come  to,  at  $3' 67  per  yard  ? 

NOTE.  —  Reduce  the  common  fractions  to  decimals. 
Do  the  same  wherever  they  occur  in  the  examples  which 
°4      "       !  follow. 
3^0-    «      }  Ans.  36'475  yards.     $133*863 +,  cost. 

2.  From  a  piece  of  cloth,  containing  36  J  yards,  a  merchant 
sold,  at  one  time,  7T^0  yards,  and,  at  another  time,  12|  yards ; 
how  much  of  the  cloth  had  he  left?  A_?is.  16'7  yds. 

3.  A  farmer  bought  7  yards  of  broadcloth  for  $33-f£,  two 
barrels  of  flour  for  $14T\,  three  casks  of  lime  for  S7|-,  and  7 
pounds  of  rice  for  $f ;  what  was  the  cost  of  the  whole  ? 

NOTE:  —  The  following1  examples  are  to  be  performed  according  to 
the  rule  in  Tf  77,  or  in  «fi  83,  or  in  ^[  85. 

4.  At  121  cents  per  lb.,  what  will  37J  Ibs.  of  butter  cost? 

Ans.  $4<7182. 


128  DECIMAL  FRACTIONS.  H  100> 

5    At  $17'37  per  ton  for  hay,  what  will  llf  tons  cost? 

Am.  $201'92f. 

6.  The  above  example  reversed.     At  $201<92f  for  11  f  tons 
of  hay,  what  is  that  per  ton  ?  Ans.  $17'37. 

7.  If  '45  of  a  ton  of  hay  cost  $9,  what  is  that  per  ton  ? 

Ans.  $20. 

8.  At  <4  of  a  dollar  a  gallon,  what  will  '25  of  a  gallon  of 
molasses  cost?  Ans.  $'1. 

9.  What  will  2300  Ibs.  of  hay  come  to,  at  7  mills  per  Ib.  ? 

Ans.  $16' 10. 

10.  What  will  765|  Ibs.  of  coffee  come  to,  at  18  cents  per 
Ib.  ?  Am.  $137<79. 

11.  Bought  23  firkins  of  butter,  each  containing  42  pounds, 
for  16 \  cents  a  pound ;  what  would  that  be  a  firkin  ?  and  how 
much  for  the  whole  ?  Am.  $159'39  for  the  whole. 

12.  A  man  killed  a  beef,  which  he  sold  as  follows,  viz., 
the  hind  quarters,  weighing  129  pounds  each,  for  5  cents  a 
pound ;  the  fore  quarters,  one  weighing  123  pounds,  and  the 
other  125  pounds,  for  4|  cents  a  pound ;  the  hide  and  tallow, 
weighing  163  pounds,  for  7  cents  a  pound ;  to  what  did  the 
whole  amount  ?  Am.  $35'47. 

13.  A  farmer  bought  25  pounds  of  clover  seed  at  11  cents 
a  pound,  3  pecks  of  herds  grass  seed  for  $2'25,  a  barrel  of 
flour  for  $6'50,  13  pounds  of  sugar  at  12J  cents  a  pound;  for 
which  he  paid  3  cheeses,  each  weighing  27  pounds,  at  8J  cents 
a  pound,  and  5  barrels  of  cider  at  $1'25  a  barrel.     The  bal 
ance  between  the  articles  bought  and  sold  is  1  cent;  is  it  for 
or  against  the  farmer  ? 

14.  A  man  dies,  leaving  an  estate  of  $71600 ;  there  are 
demands  against  the  estate,  amounting  to   $39S76'74;  the 
residue  is  to  be  divided  between  7  sons ;  what  will  each  one 
receive  ?  Ans.  $4531'894f . 

15.  How  much  coffee,  at  25  cents  a  pound,  may  be  had  for 
100  bushels  of  rye,  at  87  cents  a  bushel  ?  Ans.  348  pounds. 

16.  At  12J  cents  a  pound,  what  must  be  paid  for  3  boxes 
of  sugar,  each  containing  126  pounds  ?  Ans.  $47'25. 

17.  If  650  men  receive  $86'75  each,  what  will  they  all 
receive  ?  Ans.  $56387<50. 

18.  A  merchant  sold  275  pounds  of  iron  at  6|  cents  a 
pound,  and  took  his  pay  in  oats,  at  $0'50  a  busnel;  how 
many  bushels  did  he  receive  ?  Ans.  34'375  bushels. 

19.  How  many  yards  of  cloth,  at  $4'66  a  yard,  must  be 
given  for  18  barrels  of  flour,  at  $9'32  a  barrel  ? 

Ans.  36  yards. 


IT  101.  BILLS.  129 

20.  What  is  the  price  of  three  pieces  of  cloth,  the  first  con 
taining  16  yards,  at  $3'75  a  yard ;  the  second,  21  yards,  at 
$4'50  a  yard  ;  and  the  third,  35  yards,  at  $5'12J  a  yard  ? 

Ans.  $333'87J. 


BILLS. 

IT  1O1.  A  Bill,  in  business  transactions,  is  a  written  list 
ot  the  articles  bought  or  sold,  and  their  prices,  together  with 
the  entire  cost  or  amount  cast  up. 

No.  1.  —  Bill  of  Sale.     Payment  received. 

Boston,  May  25th,  1847. 
James  Brown,  Esq. 

Bought  of  Hastings  &;  Belding, 
6  yards  black  broadcloth,  ®         $3'00 

2|    "     cambric,  " 

2  dozen  buttons,  "  '15 

4  skeins  sewing  silk,  '04 

25  Ibs.  brown  sugar,  "  '09 

Received  payment,  $21 '06. 

Hastings  &  Belding. 

No.  2. — Bill  of  Sale.     Charged  in  account. 

New  Orleans,  Aug.  1st,  1847. 
Gen.  Z.  Taylor, 

To  Daniels  &  Thomas,  Dr. 
To  278  bbls.  beef,  <®        $9'75 

"  191     "     pork,  "         12'00 

"  250     "     flour,  "  5'70J 

"  500  sacks  Indian  meal.  "  '62i 


Charged  in  acc't.  Amount,  $6741 ' 

Daniels  &  Thomas. 

No.  3.— Barter  Bill. 

Buffalo,  Sept.  15th,  1847. 
Mr.  D.  F.  Standart, 

To  O.  B.  Hopkins  &  Co.,  Dr. 
To  15  Ibs.  brown  sugar,  <a        $  '10 

"     2    «    Y.  H.  tea,  "  '87J 

"  24    "    mackerel,  "  '04J 

"     3  gal.  molasses,  "  '42 

"   16  yds.  sheeting,  "  '09 

IT" 


130  BILLS.  H  101. 

Cr. 

By   4  doz.  eggs,  0       $  '08 

"     8  Ibs.  butter,  "  '14 

"  40    «    cheese,  "  <07£ 

"  note  at  30  days,  to  balance,  2<59 


$7<03 

O.  B.  Hopkins  &  Co. 
by  L.  D.  Swift. 

No.  4. — BUI  of  goods  sold  at  wholesale. 

New-York,  April  5th,  1847. 
Davis  &;  Horton, 

Bought  of  Barnes  Porter  &  Co. 
3  hhds.  molasses,  118  gal.  eacn,  <®     $  '31 

2  "      brown  sugar,  975  aud.  850  Ibs.    "         '09J 

3  casks  rice,  205  Ibs.  each.  "         <04J 
5  sacks  coffee,  75  "    - :;  "         '11 
1  chest  H.  tea,  86  "      "                           "         '92 


$43146 
Rec'd  payment,  by  note,  at  60  days. 

For  Barnes  Porter  &  Co. 

James  D.  Willard. 

It  is  sometimes  practised,  in  collecting  and  settling  accounts, 
to  make  a  copy  of  each  individual  account,  and  present  it  to 
the  person  for  his  inspection. 

No.  5.  —  Copy  of  an  individual  account. 
Frank  H.  Wright, 

In  acc't  with  Edward  F.  Cooper, 
1847.  Dr. 

Jan.  7.     To  125  bushels  corn,  &     $  '50 

"     "        "     20       "       apples,  "          '31 

March  13.  "     12       "       rye,  "          '62 

"      20,  "     15|  Ibs.  cast  steel,  "          '24 


Questions.  —  IT  10 1.  What  is  a  bill  ?  If  the  amount  of  the  bill  be 
paid  at  the  time,  how  is  it  shown?  Which  bill  is  an  example  of  this? 
if  charged  in  account,  how  is  it  shown  ?  example  ?  How  does  a  baiter 
bill  differ  from  a  bill  of  sale  ?  In  what  order  are  the  articles  bought  and 
sold  arranged?  What  is  practised  in  collecting  and  settling  accounts? 
How  does  such  a  copy  differ  from  a  barter  bill  ?  To  which  of  the  bill* 
must  the  bill  to  be  made  out  conform  ?  and  what  will  it  be  called  ? 


1T102.  COMPOUND  NUMBERS.                               131 

1847.  Cr. 

Feb.  15.  By  3  cows,                             <a>    $17'00 

"     22.  "  5  sheep,                             "         2'50 


Amount  due  me,  $16,42 

Edward  F.  Cooper. 
Baltimore,  May  9th,  1847. 

The  pupil  is  required  to  make  out  a  bill  from  the  statement 
contained  in  the  following  example. 

Wm.  Prentiss  sold  to  David  S.  Platt  780  Ibs.  of  pork,  at  6 
cents  per  Ib. ;  250  Ibs.  of  cheese,  at  8  cents  per  Ib. ;  and  154 
Ibs.  of  butter,  at  15  cents  per  Ib. ;  in  pay  he  received  60  Ibs. 
of  sugar,  at  10  cents  per  Ib.  ;  15  gallons  of  molasses,  at  42 
cents  per  gallon  ;  J  barrel  of  mackerel,  $3' 75  ;  4  bushels  of 
salt,  at  $T25  per  bushel ;  and  the  balance  in  money  :  how 
much  money  did  he  receive  ?  Ans.  $68'85. 


COMPOUND   NUMBERS. 

IT  1O9.  When  several  abstract  numbers,  or  several  de 
nominate  numbers  of  the  same  unit  value,  are  employed  in  an 
arithmetical  calculation,  they  are  called  simple  numbers,  and 
operations  with  such  numbers  are  called  operations  in  simple 
numbers.  Thus,  if  it  were  required  to  add  together  7  gallons, 
9  gallons,  and  5  gallons,  the  numbers  are  simple  numbers, 
being  denominate  numbers  of  the  same  unit  value,  (1  gal.,) 
and  the  operation  is  an  addition  of  simple  numbers.  We 
have  had,  also,  subtraction,  multiplication,  and  division  of 
simple  numbers. 

But  when  several  numbers  of  different  unit  values  are  em 
ployed  to  express  one  quantity,  the  whole  together  is  called  a 
compound  number.  Thus,  12  rods,  9  yards,  2  feet,  6  inches, 
•employed  to  express  the  length  of  a  field,  is  a  compound  num 
ber.  So  also,  9  gallons,  2  quarts,  1  pint,  employed  to  express 
a  quantity  of  water,  is  a  compound  number. 

NOTE.  —  The  word  denomination  is  used  in  compound  numbers  to 


Questions.  —  ^[  102.  What  are  simple  numbers?  Examples. 
What  are  operations  in  such  numbers  called  ?  What  is  a  compound 
number  ?  Give  examples  other  than  those  in  the  book  What  is  meant 
by  the  word  denomination  ? 


132  COMPOUND  NUMBERS.  If  J03,  104 

denote  the  name  of  the  unit  considered.  Thus,  bushel  and  peck  are 
names  or  denominations  of  measure  ;  hour,  minute  and  second  are  de 
nominations  of  time. 

IT  1O3.  The  fundamental  operations  of  addition,  subtrac< 
tion,  multiplication  and  division,  cannot  be  performed  on  com 
pound  numbers  till  we  are  acquainted  with  the  method  of 
changing  numbers  of  one  denomination  to  another  without 
altering  their  value,  which  is  called  Reduction.  Thus,  we 
wish  to  add  2  bushels  3  pecks,  and  3  bushels  1  peck,  together. 
They  will  not  make  9  bushels  nor  9  pecks,  (adding  together 
the  several  numbers,)  since  some  of  the  numbers  express 
bushels,  and  some  express  pecks.  But  2  bushels  equal  3 
pecks,  (2  times  4  pecks,  the  number  of  pecks  in  a  bushel,) 
and  3  pecks  added  make  11  pecks ;  3  bushels  equal  12  pecks, 
and  1  peck  added  make  13  pecks.  Then,  11  pecks  -j-  13 
pecks  =  24  pecks.  Hence,  before  proceeding  further,  we 
must  attend  to  the 

Reduction  of  Compound  Numbers. 

STERLING  OR  ENGLISH  MONEY. 

<fF  104.  Money  is  expressed  in  different  denominations, 
and  4  dollars,  3  dimes,  7  cents,  5  mills  =  $4'375,  employed 
to  express  one  sum  in  Federal  money  is  a  compound  number. 
But  as  the  denominations  in  Federal  money  vary  uniformly 
in  a  tenfold  proportion,  (IT  93,}  being  conformed  to  the  Arabic 
notation  of  whole  numbers,  the  operations  in  it  are  as  in  whole 
numbers. 

The  denominations  in  English  (called,  also,  sterling)  money, 
pounds,  shillings,  pence  and  farthings,  do  not  vary  uniformly, 
but  according  to  the  following 

TABLE. 

NOTE  1.  —  All  the  tables  in  Reduction  of  Compound  Numbers 
must  be  carefully  committed  to  memory  by  the  pupil. 

4  farthings  (qrs.)  make  1  penny,  marked  d. 
12  pence  (plural  of  penny)  1  shilling,  "  s. 
20  shillings  1  pound,  "  £. 

NOTE  2.  —  Farthings  are  often  written  as  the  fraction  of  a  penny, 
thus,  1  farthing  =  id.,  2  farthings  =  £d.,  3  farthings  =  |d. 

Questions.  —  ^[  103.  What  is  reduction  ?  Whence  its  necessity  ? 
Explain  by  the  example  of  adding  bushels  and  pecks.  To  what,  then, 
must  we  attend  before  proceeding  further? 


IT  104. 


COMPOUND  NUMBERS. 


133 


NOTE  3. — The  value  of  these  denominations  in  Federal  money  is 
nearly  as  follows : 

Iqr.  =  £f  £  °f *  cent 

Id.  =         2^  cents. 

Is.  •  ==       24|  cents. 

1£.  =  S4'84 

4s.   Id.  2^-rnrs.  =S1'00 


There  is  in  England 
a  gold  coin,  called  a 
sovereign,  the  value 
of  which  is  £\. 


1.    How  many  farthings  in 
5  pence  ? 

1ST  OPERATION.        SOLUTION.  — 

4  We  may  multi- 
§  ply  the  number 

of  farthings  (4) 

in    1    penny   by 

20?r5'         the   number    of 

2D  OPERATION.     pence?  (5.)      (^[ 

5  46.)     Or,  as  ei- 
4  ther  factor  may 

be     made     the 

2Qars          multiplicand.  (1 
21,)     we     may 

multiply  the  number  of  pence  (5) 
by  the  number  of  farthings  in  1 
penny.  Ans.  20qrs. 

3.    How  many  farthings  in 

3  pence  ? 6  peace  ? 

9  pence  ? 7  pence  ? 

2  pence? 10  pence  ? 


12  pence = 


11  pence  ? 
1  shilling? 

5.    How  many  pence  in  3 

shillings?   5  shillings? 

5S.  8d.  ?  7s.  ?  - — 

8s.  4d.?  12s.?  15s. 

6d.  f 

7.    How  many  shillings  in 
£3  ?  £5  ?  . £4  2S.  ? 

£6  11s.? 


2.    In    20    farthings,   how 
many  pence  ? 

SOLUTION.  —  We 
OPERATION,  have  given  the  num- 
4 )  20         ber  of  farthings  in  1 
—        penny    to    find    the 
5^    number  of  pence  in  a 
given  number  of  far 
things,  (20,)  and  we  divide  the 
number  of  farthings  in  the  num 
ber  of  pence  by  the  number  in  1 
penny,  (^[46.)  Ans.  5d. 


4.  How  many  pence  in  12 

farthings  ?  24  farthings  ? 

36  farthings?  28 

farthings  ?  8  farthings  ? 

40  farthings  ?  -  -  44 

farthings  ?  48  farthings  ? 

6.  How  many  shillings  in 

36  pence  ?  60  pence  ? 

68d.  ?  84d.  ?  

lOOd.?  144d.?  

188d.  ? 

8.  How  many  pounds  in 

60s.?  lOOs/?  82s.? 

131s.  ? 


Questions.  —  U  1O4.  What  is  said  of  operations  in  Federal  money  ? 
Wha.  are  the  denominations  of  English  money?  the  signs?  How 
do  they  vary  differently  from  those  of  Federal  money?  Give  the  table. 
How  are  farthings  written?  What  is  the  value  of  a  pound  sterling  in 
Federal  money?  Explain  the  first  operation  of  Ex.  1 :  the  second  oper 
ation  Explain  Ex.  2.  Of  how  many  kinds  is  reduction?  what  are 
they  What  is  reduction  descending  ?  —  reduction  ascending  ? 
12 


134 


COMPOUND  NUMBERS. 


H105 


The  changing  of  higher 
denominations  to  lower,  as 
pounds  to  shillings,  is  called 
Reduction  Descending,  and  is 
performed  by  multiplication. 

REDUCTION    DESCENDING. 

1T1O3.     1.   In  £17   13s. 

6|d.,  how  many  farthings  ? 

OPERATION. 
17£  13s.  Qd.  3qrs. 
20 

3535.  in  17£  13s. 
12 


4242d.  in  17£  13s.  6d. 
4 

1697] qrs.  Ans. 

In  17£  13s.  6d.  3qrs. 

SOLUTION.  —  We  multiply  17 £ 
by  20,  the  shillings  in  IX',  and 
add  in  the  13s.  to  get  the  number 
of  shillings  in  17 £  13s.,  which  is 
353.  This  number  we  multiply 
by  12,  adding  in  the  6d.  given,  to 
get  the  number  of  pence,  4242, 
which  we  multiply  by  4,  adding 
in  the  3qrs.  given,  to  get  the  num 
ber  of  qrs.  or  farthings,  which  is 
1697lqrs. 


Hence,  for  Reduction  De 
scending, 

RULE. 

Multiply  each  higher  de 
nomination  by  the  number 
which  it  takes  of  the  next  less 


The  changing  of  lower  de 
nominations  to  higher,  as  shil 
lings  to  pounds,  is  called  Re 
duction  Ascending,  and  is 
performed  by  division. 

REDUCTION    ASCLNDUNG. 

2.  In  16971  farthings,  how 
many  pounds  ? 


OPERATION. 


17£  13s. 
Ans.  £17  13s.  6d.  3grs. 


SOLUTION.  —  We  divide  the 
whole  number  of  farthings  by  4, 
the  number  in  Id.,  to  get  the 
number  of  pence ;  for  as  many 
times  as  4  can  he  subtracted  from 
16971,  so  many  pence  there  will 
be,  which  is  4242d.  and  3qrs.  re 
maining.  On  the  same  principle, 
dividing  the  4242  by  12,  the  quo 
tient,  353,  is  shillings,  and  the 
remainder,  6,  is  pence,  and  divid 
ing  353s.  by  20,  the  quotient,  17, 
is  pounds,  and  the  remainder,  13, 
is  shillings. 

Hence,  for  Reduction  As 
cending, 

RULE. 

Divide  each  lower  denomi 
nation  by  the  number  which 
it  takes  of  it  to  make  one  of 


Questions.  — TI 1O5.     Explain  the  first  example.     Give  the  rule 
for  reductic  n  descending.  Ex.  2.     Give  the  rule  for  reduction  ascending. 


f  106.  COMPOUND  NUMBERS.  135 

to  make  1  of  this  higher,  in-  the  next  higher.     Proceed  in 

creasing  the  product  by  the  this  way  till  the  work  is  done. 

given  number,  if  any,  of  this 

lower  denomination.    Proceed 

in  this  way  till  the  work  is 

done. 

EXAMPLES   FOR    PRACTICE. 

3.  Reduce  £32  15s.  Sd.  to  4.  Reduce  31472  farthings 

qrs.  to  pounds. 

5.  Reduce  £7  14s.  6d.  1  6.  Reduce  7417  qrs  to 

or.  to  qrs.  pounds. 

7.  In  £91  11s.  3Jd.,  how  8.  In  87902  farthings,  now 

many  farthings  ?  many  pounds  ? 

9.  In  £40  12s.  8d.,  how  10.  In  9752  pence,  how 

many  pence  ?  many  pounds  ? 

11.  In  £1  18s.  4Jd.,  how  12.  In  921  half  pence,  how 

many  half  pence  ?  many  pounds  ? 


Weight. 

I.   AVOIRDUPOIS  WEIGHT. 

IF  1O6.  Avoirdupois  Weight  is  employed  in  all  the  ordi 
nary  purposes  of  weighing.  The  denominations  are  ton^ 
pounds,  ounces,  and  drams. 

TABLE. 

16  drams  (drs.)  make  1  ounce,  marked  oz. 

16  ounces  "      1  pound,  "          Ib. 

2000  pounds  "      1  ton,  "          T. 

Or,  as  was  formerly  reckoned, 

28  Ibs.  1  quarter,  "        qr. 

4  qrs.  (=112  Ibs.)     1  hundred  weight,     "      cwt. 

SOcwt.  (=2240  Ibs.)  Iton,  «         T. 

By  the  last  talle,  2240  Ibs.  make  1  ton,  which  is  sometimes 
called  the  "  long  ton  ;"  while  the  ton  of  2000  Ibs.  is  called 
the  "  short  ton."  The  long  ton  is  still  used  in  the  U.  S.  cus- 

Questions,  —  ^[  106.  What  is  the  use  of  avoirdupois  weight  ?  the 
denominations  ?  the  signs  ?  Repeat  the  table ;  the  table  by  the  old 
method.  Explain  the  difference  between  the  long  and  short  ton.  "When 
's  the  long  ten  used?  the  short  ton? 


136  COMPOUND  NUMBERS.  IT  107 

torn-house  operations,  in  invoices  of  English  goods,  and  of 
coal  from  the  Pennsylvania  mines.  But  in  selling  coal  in 
cities,  and  in  other  transactions,  unless  otherwise  stipulated, 
2000  Ibs.  are  called  a  ton. 

EXAMPLES    FOR    PRACTICE. 

1.  In  14  tons  607  Ibs.  6  oz.  2.  In  7323500  drams,  how 
12  drs.,  how  many  drams  ?  many  tons? 

SOLUTION.  —  As  there  are  2000  SOLUTION. — Dividing  the  drams 

Ibs.  in  a  ton,  we  multiply  14  by  by  16,  the  number  in  an  oz.,  the 

2000,  to  get  14  tons  to  Ibs.,  and  quotient  is  oz.  and  the  remainder 

acid  the  607  Ibs.  to  the  product,  drs.     Dividing  the  oz.  by  16,  the 

The  Ibs.  we  multiply  by  16  to  get  quotient  is  Ibs.  and  the  remainder 

them  to  oz.,  adding  in  6  oz.,  and  oz.,  and  dividing  the  Ibs.  by  2000, 

the  oz.  by  16,  adding  in  12,  and  the  quotient  is  tons  and  the  re- 

the  whole  are  in  drams.  mainder  Ibs. 

3.  In  7  tons  665  Ibs.  of  su-  4.  In  14665  Ibs.  of  sugar, 
gar,  how  many  Ibs.  ?  how  many  tons  ? 

5.  In  12  T.  15  cwt.  1  qr.  19         6.  In  7323500  drams,  how 
Ibs.  6  oz.  12  drs.  of  glass,  re-     many  tons  ? 
ceived  from  an  English  house, 
how  many  drams  ? 

7.  Received  from  Binning-         8.    In    470    packages    of 
ham.  England,  5  T.  9  cwt.  12     screws,    each    containing   26 
Ibs.   of  iron  screws,  in  pack-     Ibs.,  how  many  tons  ? 
ages  of  26   Ibs.   each  ;    how 
many  were  the  packages? 

II.    TROY  WEIGHT. 

^T  1OT.  Troy  Weight  is  used  where  great  accuracy  is 
required,  MS  in  weighing  gold,  silver,  and  jewels.  The  de 
nominations  are  pounds,  ounces,  pennyweights,  and  grains. 

TABLE. 

24  grains  (grs.)  make  1  pennyweight,  marked  pwt. 
20  pwts.  1  ounce,  "  oz. 

12  oz.  1  pound,  "  Ih. 

NOTE.  —  A  Ib.  Troy  =  5760  grs.,  and  1  Ib.  avoirdupois  =  7000 
grs.  Troy.  Hence  a  quantity  expressed  in  one  weight,  may  be 
chunged  to  the  denominations  of  the  other. 

Questions.  —  T[  1O7.  For  what  is  Troy  weight  used?  What  are 
the  denominations  ?  —  the  signs  ?  Repeat  the  table.  What  difference 
between  the  pound  Troy  and  the  pound  avoirdupois  ? 


U  108.  COMPOUND   NUMBERS.  137 

EXAMPLES    FOR    PRACTICE. 

1.  In  210  Ibs.  8  oz.  12  2.  In  50572  pwts.,  how 
pwts.,  how  many  pwts.  ?  many  Ibs.  ? 

SOLUTION.  —  Multiply  the  Ibs.  SOLUTION. — Dividing  the  pwts. 

by  12,  adding  the   8  oz.   to   the  by  20,  the  quotient  is  oz.,  and  the 

product,    and    the    sum    is    oz.,  remainder  pwts.,  and  dividing  the 

which,  multiplying  by  20,  adding  oz.  by  12,  the  quotient  is  Ibs., 

in  the  12  pwts.,  the  sum  is  pwts.  and  the  remainder  oz. 

3.    In  7  Ibs.  11  oz.  3  pwts.  4.    In  45681  grains  of  sil- 

9  grs.  of  silver,   how   many  ver,  how  many  Ibs.  ? 
grains  ? 

5.    Reduce  11  oz.  13  pwts.  6.    Reduce    5605    grs.    of 

13  grs.  of  gold  to  grains.  gold  to  ounces. 

7.    Reduce  28  Ibs.  avoirdu-  8.    Reduce  34  Ibs.  6  pwts. 

pois  to  the  denominations  of  16  grs.  Troy  to  Ibs.  avoirdu- 

Troy  weight  ?  pois.     (Consult  Note.) 

III.   APOTHECARIES'  WEIGHT. 

5T  1O8.9  Apothecaries'  Weight  is  used  by  apothecaries 
and  physicians,  in  mixing  and  preparing  medicines.  But 
medicines  are  bought  and  sold  by  avoirdupois  weight. 

The  denominations  are  pounds,  ounces,  drams,  scruples, 
and  grains. 

TABLE. 

20  grains  (qrs.)  make  1  scruple,  marked  9. 

3  9  1  dram,  "        5. 

85  1  ounce,  "        §. 

12  %  1  pound,          "       ft,. 

NOTE. — The  pound  and  ounce,  Apothecaries'  and  Troy  weight, 
are  the  same,  but  the  ounce  is  differently  divided. 

EXAMPLES    FOR    PRACTICE. 

1.  In  91b.  8§.  15.  29.  19  2.  Reduce  55799  grs.  to 
grs.  how  many  grains  ?  Ibs 

Questions. — H  1O8.  To  what  is  apothecaries'  weight  limited? 
the  denominations  ?  Give  the  table.  Make  the  sign  for  each  denomi 
nation.  What  is  s?id  of  the  pound  and  ounce? 

12* 


138  COMPOUND  NUMBERS.  IF  109. 


Measures  of  Extension. 

Extension  has  three  dimensions,  length,  breadth,  and  thick 
ness. 

I.    LINEAR  MEASURE. 

^F  1OO.  Linear  Measure  (the  measure  of  lines)  is  used 
when  only  one  dimension  is  considered,  which  may  be  either 
the  length,  breadth,  or  thickness. 

The  usual  denominations  are  miles,  furlongs,  rods,  yards, 
feet,  inches,  and  barley-corns. 

TABLE. 

3  barley-corns  (bar.)  make  1  inch,  marked  in. 

12  inches  1  foot,  "  ft. 

3  ft.  1  yard,        "       yd. 

5|  yards,  or  16J  ft.,  1  rod,  "  rd. 

40  rods  1  furlong,  "  fur. 

8  furlongs,  or  320  rods,  1  mile,        "       mi. 


69£  common  miles,        1  degree,  deg.,  or  ° ,\° 

3  geographical  mileS,    1   league,   L.,    \  used  in  measuring  distance*  at  sea. 

60  geographical  miles,  1  degree  of  latitude. 
6  feet,  1  fathom,  in  measuring  depths  at  sea. 

NOTE.  — The  geographical  mile,  used  in  measuring  latitude,  is  not 
quite  uniform,  but  is  always  a  little  less  than  1^-  common  miles,  as 
the  degree  varies  from  68:]  to  69£  miles. 

The  degree  of  longitude  grows  shorter  towards  the  poles,  where  it 
is  nothing.  Instead  of  the  barley-corn,  inches  are  now  divided  into 
eighths  and  tenths. 

EXAMPLES    FOR    PRACTICE. 

1.  How  many  inches  in  the  2.    In  1577664000  inches, 

equatorial    circumference    of  how  many  miles  ?    How  many 

the  earth,    it  being  360   de-  degrees  of  the  equatorial  cir- 

grees  ?  cumference  ? 

Questions.  —  If  1O9.  How  many  dimensions  has  extension? 
What  are  they  ?  What  is  linear  measure  ?  Give  the  denominations, 
and  the  sign  of  each.  Repeat  the  table.  How  is  the  inch  usually 
divided?  For  what  is  the  fathom  used?  For  what  is  the  geographical 
mile  used  ?  —  its  length  ?  What  is  said  of  the  degree  of  longitude,  and 
its  length  ?  —  of  a  degree  of  latitude  ?  What  causes  the  difference  in  the 
length  of  degrees?  Fcr  what  is  the  league  used,  and  about  what  is  its 
'ength  in  common  mile j  ? 


1T110. 


COMPOUND  NUMBERS. 


139 


3.  How  many  inches  from 
Boston  to  Washington,  it  be 
ing  482  miles  ? 

5.  How  many  times  will  a 
wheel,  16  feet  6  inches  in  cir 
cumference,  turn  round  in  go 
ing  from  Boston  to  Providence, 
it  being  40  miles  ? 

7.  If  a  man  step  2  feet  6 
inches  at  once,  how  many 
steps  will  he  take  in  walking 
43  miles  ? 


4.    In    30539520 
how  many  miles  ? 


inches, 


6.  If  a  wheel,  16  feet  6 
inches  in  circumference,  turn 
round  12800  times  in  going 
from  Boston  to  Providence, 
what  is  the  distance  ? 

8.  A  man  walked  90816 
steps,  of  2  feet  6  inches  each, 
in  a  day;  how  many  miles 
did  he  walk? 


CLOTH  MEASURE. 


If  HO.  Cloth  Measure  is  a  species  of  linear  measure, 
being  used  to  measure  cloth  and  other  goods  sold  by  the  yard 
in  length,  without  regard  to  the  width. 

The  denominations  are  ells,  yards,  quarters,  and  nails. 

TABLE. 

4  nails,  (na.,)  or  9  inches,  make  1  quarter,   marked    qr. 

4  qrs.,  or  36  inches,  1  yard,             "         yd. 
3qrs.  1  ell  Flemish,"    E.F1. 

5  qrs.  1  ell  English,  "     E.E. 

6  qrs.  1  ell  French,  "    E.Fr. 

NOTE.  —  Eighths  and  sixteenths  of  a  yard  are  now  used  instead  of 
nails. 

EXAMPLES    FOR    PRACTICE. 


1.  In  573  yds.  1  qr.  1  na., 
how  many  nails  ? 

3.  In  296  E.E.  3  qrs.,  how 
many  nails  ? 

5.  In  151  E.E.,  how  many 
yards  ? 

7.  In  29  pieces  of  cloth, 
each  containing  36  E.  FL, 
how  many  yards  ? 


2.  In  9 173  nails,  how  many 
yards  ? 

4.  In  5932  nails,  how  many 
E.E.? 

6.  In  188  yds.  3  qrs.,  how 
many  E.  E.  ? 

8.  In  783  yds.,  how  many 
E.  Fl.  ? 


Questions.  —  If  1 10.  "What  is  cloth  measure  ?  How  used  ?  Why 
a  species  of  linrar  measure  ?  Give  the  denominations,  and  the  sign  of 
each  ;  the  table.  What  is  used  instead  of  nails?  How  may  yards  be 
reduced  to  E.  E.  ?  to  E.  Fr.  ?  to  E.  Fl.  ?  How  each  of  these  to  yards ? 


140 


COMPOUND   NUMBERS. 


fill. 


II,    LAND  OR  SQUARE  MEASURE. 

IT  111.     Square  Measure  is  used  in  measuring  land,  and 
other  things  wherein  length  and  breadth  are  considered. 

NOTE. — It  takes  3  feet  in  length  to  make  1 
linear  yard. 

But  it  requires  a  square,  3  feet  =  1  linear 
yard  in  length,  and  3  feet  =  1  linear  yard  in 
breadth,  to  make  1  square  yard.  3  feet  in 
length  and  1  foot  in  width,  make  3  square 
feet,  (3  squares  in  a  row,  ^f  48.)  3  feet  in 
length  and  2  feet  in  width  make  3X2  =  6 
square  feet,  (2  rows  of  squares,  *J[48.)  3  feet 
in  length  and  3  feet  in  width  make  3X3  =  9 
square  feet,  (3  rows  of  squares.) 
9  sq.  ft.  =  i  sq.  yd.  It  is  plain,  also,  that  1  square  foot,  that  is,  a 
square  12  inches  in  length  and  12  inches  in 

breadth,  must  contain  12  X  12=  144  square  inches,  (12  rows,  of  12 
squares  each.) 

The   denominations  of  square  measure   are  miles,  acres, 
roods,  rods  or  poles,  yards,  feet,  and  inches. 


1  linear  yard. 

i 
. 

*U 

ft 

II 

1 

CO 

I    '    2    i    3    '• 

3  feet  =  1  yard. 

TABLE. 

144  square  inches  (sq.  in.)  make  1  square  foot,  marked  sq.  ft. 

9  square  feet  1  square  yard,       "     sq.  yd. 

30^  sq.  yds.  =  5^  X  5|,or  )  (  1  sq.  rod,  perch,  )  "    sq.  rd. 

272i  sq.  ft.  =  16J  x  16J,      {  {         or  pole,         \ 

40  square  rods  1  rood,                    " 

4  roods,  or  160  square  rods  1  acre,                    " 

640  acres  1  square  mile,        " 


P. 
R. 

A. 

M. 


EXAMPLES    FOR    PRACTICE. 


1.  In  17  acres  3  roods  12 
poles,  how  many  square  feet? 

3.  Reduce  64  square  miles 
to  square  feet. 

5.  There  is  a  town  6  miles 
square ;  how  many  square 
miles  in  that  town  ?  how  many 
acres  ? 


2.  In  776457  square  feet, 
how  many  acres  ? 

4.  In  1,784,217,600  square 
feet,  how  many  square  miles  ? 

6.  Reduce  23040  acres  to 
square  miles. 


Questions.  —  If  111.  What  is  square  measure?  "What  is  a 
square?  Draw  or  describe  a  square  inch;  a  square  yard;  a  figure 
snowing  the  square  inches  contained  in  a  square  foot.  How  many 
square  inches  in  a  row,  and  how  many  rows  of  square  inches  would  it 
contain?  Multiply  questions  at  pleasure.  What  are  the  denominations 
of  square  measure?  Repeat  the  table.  How  does  square  measure 
differ  from  linear  measure  ? 


1T  112,  113.  COMPOUND  NUMBERS.  141 

7.    How  many  square  feet  8.    In    5510528179200000 

on  the   surface  of  the  globe,  square  feet,  how  many  square 

supposing-  it  contain  197,663,-  miles  ? 
000  square  miles  ? 

IT  112.     The    Surveyor's,   or   what   is   called    Gunter's 
Chain,  is  generally  used  in  surveying  land. 
It  is  4  rods,  or  66  feet  in  length,  and  consists  of  100  links. 

TABLE    FOR   LINEAR    MEASURE. 

7T9^  inches  make  1  link,  marked  1. 
25  links  1  rod,  "  rd. 

4  rods,  or  66  feet,  1  chain,  "  C. 
SO  chains  1  mile,  "  mi. 

1.    In  5  mi.   71  C.,   how  2.    In    471     chains,    how 

many  chains  ?  many  miles  ? 

.  3.    Reduce  2  mi.    15  C.   3  4.    Reduce  17593  links  to 

rds.  18  1.  to  links.  miles. 

5.    In   75  C.,   how  many  6.   In  4950  feet,  how  many 

feet  ?  chains  ? 

TABLE  FOR  SQUARE  MEASURE. 

625  square  Hnks  <„.  1.)  malce  j  {^"J  j  ™™  *«• 

16  square  poles  1  square  chain,      "         sq.  C. 

10  square  chains  1  acre,  "  A. 

NOTE.  — Land  is  generally  estimated  in  square  miles,  acres,  roods 
uare  poles  or  perches. 


iiid  &([ 


7.    Reduce  8  A.  2  sq.  C.  7  8.    In  824831   sq.  L,   how 

P.      456     sq.  1.     to    square  many  acres  ? 
links. 

9.    In   80   A.,  how   many  10.   In  8000000  sq.  L,  how 

square   chains  ?    how   many  many  square  chains  ?    In  800 

square  links?  sq.  C.,  how  many  acres? 

III.   CUBIC  OR  SOLID  MEASURE. 

IT  113.  Cubic  or  Solid  Measure  is  used  in  measuring 
things  that  have  length,  breadth,  and  thickness;  such  as  tim 
ber,  wood,  earth,  stone,  &c. 

Questions.  —  IT  112.  What  is  generally  used  in  surveying  land? 
What  is  its  length  ?  Of  how  many  links  does  it  consist  ?  Repeat  the 
table  for  linear  measure ;  for  square  measure.  How  is  land  generally 
estimated  ? 


142 


COMPOUND  NUMBERS. 


If  113. 


NOTE  1.  — It  has  been  shown  (^f  111,)  that  1  square  yard  contains 
3X3  =  9  square  feet. 

A  block  3  feet  long,  3  feet 
wide  and  3  feet  thick,  is  a 
cubic  yard.  The  accompanying 
figure  represents  such  a  block. 

Were  a  portion  1  foot  in  thick 
ness  cut  off  from  the  top  of  this 
block,  the  part  cut  off  would  be 
3  feet  long,  3  feet  wide,  and  1  foot 
thick,  and  would  contain  3x3 
X  1  =  9  cubic  feet. 
The  bottom  part  being  3  feet  long,  3  feet  wide,  and  2  feet 
thick,  would  contain-3  X  3  X  2  =  18  cubic  feet. 

But  the  entire  block  being  3  feet  long,  3  feet  wide,  and  3 
feet  thick,  contains  3  X  3  X  3  =  27  cubic  feet. 

It  is  plain  also,  that  a  cubic  foot,  that  is,  a  solid  body,  12 
inches  long,  12  inches  wide,  and  12  inches  thick,  will  contain 
12  X  12  X  12=  1728  cubic  or  solid  inches. 

The  denominations  of  cubic  measure  are  cords,  tons,  yards, 
feet  and  inches.  -  « 


1  yd.  =  3  feet  long. 


TABLE. 

1728  cubic  inches,  (cu.  in.)  " 

=  12  X  12  X  12,  that  is, 

12  inches  in  length,  12 

•  make  1  cubic  foot,  marked  cu.  ft 

in    breadth,   and    12   in 

thickness, 

27  cubic  feet,  3  X  3  X  3,' 
50  feet  of  round  timber,  or 

1  cubic  yard,     "     c^f  yd. 

1m 

40  feet  of  hewn  timber, 

ton,                "           T. 

42  rnhiV  fppt                                      5  l  ton  of  shiPPing'  1        T 

<   Used   in   measuring  the  ca-   (             •»•  • 

\       pacity  of  ships,                         ) 

16  cubic  feet                                   j  }  *01*  f?ot'  °^   I       C.  ft. 
(  1  foot  of  wood,  } 

10Q  '                                                     1  cord  of  wood,              C. 

Questions.  —  *[[  1 13.  "What  is  cubic  measure  ?  "What  distinctions 
do  you  make  between  a  line,  a  surface  and  a  solid?  What  is  a  cube? 
a  cubic  inch  ?  a  cubic  foot  ?  a  cubic  yard  ?  For  what  is  cubic  or  solid 
measure  used?  What  are  its  denominations?  Repeat  the  table.  For 
what  is  the  cubic  ton  used  ?  What  do  you  understand  by  a  ton  of  round 
timber  ?  What  are  the  dimensions  of  a  pile  of  wood  containing  1  cord  1 
What  is  a  cord  foot  ? 


1F114.  COMPOUND  NUMBERS.  143 

NOTE  2.  —  A  cubic  ton  is  used  for  estimating  the  cartage  and  trans 
portation  of  timber.  A  ton  of  round  timber  is  such  a  quantity  (about 
50  feet)  as  will  make  40  feet  when  hewn  square. 

NOTE  3.  — A  pile  or  load  of  wood  8  feet  long,  4  feet  wide,  and  4 
feet  high,  contains  1  cord.  8  X  4  X  4  =  128  cubic  feet.  A  cord  foot 
is  1  foot  in  length  of  such  a  pile. 

1.    Reduce  9  tons  of  round  2.    In  777600  cubic  inches, 

timber  to  cubic  inches.  how  many  tons  of  round  tim 
ber? 

3.    In  37  cord  feet  of  wood,  4.    In   592    solid    feet    of 

how  many  solid  feet  ?  wood,  how  many  cord  feet  ? 

5.    Reduce  8  cords  of  wood  6.    In  64  cord  feet  of  wood, 

to  cord  feet.  how  many  cords  ? 

7.    In    16   cords   of  wood,  8.    2048  solid  feet  of  wood, 

how   many  cord    feet?    how  how  many  cord   feet?    how 

many  solid  feet  ?  many  cords  ? 

9.    In  25  C.,  5  C.  ft.,  9  cu.  10,    In      5684967      cubic 

ft.,  1575  cu.  in.  of  wood,  how  inches,  how  many  cords  ? 
many  cubic  inches  ? 


Measures  of  Capacity. 

I.   WINE  MEASURE. 

^T  114.  Wine  Measure  is  used  in  measuring  all  liquids 
except  ale,  beer,  and  milk. 

The  denominations  are  tuns,  pipes,  hogsheads,  tierces,  bar 
rels,  gallons,  quarts,  pints,  and  gills. 

TABLE. 

4  gills  (gi.)  make  1  pint,  marked    pt. 

2  pints  1  quart,  "  qt. 

4  quarts  1  gallon,  "  gal. 

31J  gallons  1  barrel,  "  bar. 

42  gallons  1  tierce,  "  tier. 

63  gallons,  or  2  barrels,  1  hogshead,  "  hhd. 

2  hogsheads  1  pipe,  "  P. 

2  pipes,  or  4  hogsheads,  1  tun,  "  T. 

NOTE.  — The  wine  gallon  contains  231  cubic  inches.  A  hogshead 
rf  molasses,  &c  ,  is  no  definite  quantity,  but  is  estimated  by  the  gal 
lon. 

Questions.  —  If  114.  For  what  is  wine  measure  used  ?  What  are 
its  denominations  ?  Repeat  the  table.  How  many  cubic  inches  iii  a 
wine  gallon  ?  How  many  gallons  in  a  hogshead  of  molasses  ?  &c 


144  COMPOUND  NUMBERS.  11115,116. 

1.  Reduce  12  pipes  of  wine  2.  In  12096  pints  of  wine 

to  pints.  how  many  pipes  ? 

3.  In  9  P.  1  hhd.  22  gals.  4.  Reduce  39032  gills  to 

3  qts.,  how  many  gills?  pipes. 

5.  In  25  tierces,  how  many  6.  In  33600  gills,  how 

gills  ?  many  tierces  ? 

II.  BEER  MEASURE. 

IT  1 15.  Beer  Measure  is  used  in  measuring  beer,  ale 
and  milk. 

The  denominations  are  hogsheads,  barrels,  gallons,  quarts, 
and  pints. 

TABLE. 

2  pints  (pts.)  make  1  quart,  marked  qt. 

4' quarts  1  gallon,  "  gal. 

36  gallons  1  barrel,  "  bar. 

54  gallons,  or  1 J  barrels,  1  hogshead,  "  hhd. 

NOTE.  — The  beer  gallon  contains  282  cubic  inches. 

1.  Reduce  47  bar.  18  gal.  2.  In  13680  pints  of  ale, 

of  ale  to  pints.  how  many  barrels  ? 

3.  In  29  hhds.  of  beer,  4.  Reduce  12528  pints  to 

how  many  pints  ?  hogsheads. 

III.  DRY  MEASURE. 

5f  116.  Dry  Measure  is  used  in  measuring  all  kinds  of 
grain,  fruit,  roots,  (such  as  carrots  and  turnips,)  salt,  charcoal, 
&c. 

The  denominations  are  chaldrons,  quarters,  bushels,  pecks, 
quarts,  and  pints. 

TABLE. 

2  pints  (pts.)  make  1  quart,  marked  qt. 

8  quarts  1  peck,         "      pk. 

4  pecks  1  bushel,      "      bu. 

8  bushels  1  quarter,     "       qr. 

36  bushels  1  chaldron,  "      ch. 

NOTE  1.  —  The  dry  gallon  contains  268-f  cubic  inches.  The 
Winchester  bushel,  which  is  adopted  as  our  standard,  contains  2150f 
cubic  inches.  It  is  18£  inches  in  diameter,  and  8  inches  deep. 

The  quarter  of  8  bushels  is  an  English  measure. 

Questions.  —  If  115.  What  is  the  use  of  beer  measure?  What 
are  its  denominations?  Repeat  the  table.  How  many  cubic  inches  in 
a  beer  gallon  ? 


1f  117.  COMPOUND  NUMBERS.  145 

NOTE  2.  —The  Imperial  gallon,  adopted  in  Great  Britain  in  1826, 
for  all  liquids  and  dry  substances,  contains  277TVcrV  cubic  inches. 

1.    In  75  bushels  of  wheat,  2.    In     4800     pints,     how 

how  many  pints  ?  many  bushels  ? 

3.    Reduce  42  chaldrons  of  4.    In    6048    pecks,    how 

coal  to  pecks.  many  chaldrons  ? 

5.    In  273  qrs.  6  bu.  3  pks.  6.    In    140223   pints,   how 

7   qts.    1  pt.  of  wheat,  how  many  quarters  ? 
many  pints  ? 


Time. 

IT  117.     Time  is  the  measure  of  duration. 
The  denominations  are  years,  months,  weeks,  days,  hours, 
minutes,  and  seconds. 

TABLE. 

60  seconds  (s.)  make  1  minute,  marked  m. 

60  minutes  1  hour,  "  h. 

24  hours  1  day,  d. 

7  days  1  week,  "  w 

52  weeks  1  day  5  hours  48  min-  j) 

utes  48  seconds,  or  365  days  5  >        1  year,  "       yr. 

hours  48  minutes  48  seconds,       ) 

NOTE  1.  —  As  there  is  nearly  i  of  a  day  more  than  365  days  in  a 
year,  we  add  1  day  to  February  of  certain  years,  thus  giving  them  366 
days.  If  the  excess  was  just  \  of  a  day,  we  would  add  1  day  to  every 
4th  year,  thus  making  the  years  average  365  days  6  hours,  the  odd 
day  being  added  to  every  year  exactly  divisible  by  4. 

But  as  the  excess  is  not  quite  6  hours,  lacking  about  |  of  a  day  in 
100  years,  a  year  divisible  by  100,  though  divisible  by  4,  has  only  365 
days,  unless  it  be  divisible  by  400,  when  it  has  366  days.  Thus, 
1844  and  1600  had  366  days  each,  but  1845  and  1700  had  only  365 
days  each. 

A  year  of  366  days  is  called  Bissextile,  or  Leap  year. 

The  calendar  months,  into  which  the  year  is  divided,  are  from  28  to 
31  days  in  length. 

Questions.  —  ^  116.  Eor  what  is  dry  measure  used  ?  What  are 
its  denominations  ?  Repeat  the  table.  How  many  cubic  inches  in  a  dry 
gallon?  Describe  the  Winchester  bushel.  What  measure  is  the  quar 
ter?  What  is  said  of  the  Imperial  gallon?  Which  is  the  larger  quan 
tity,  a  quart  of  milk,  or  a  quart  of  salt  ?  —  a  quart  of  milk,  or  a  quart  of 
vinegar  ?  —  a  quart  of  oats,  or  a  quart  of  cider  ? 
13 


146 


COMPOUND  NUMBERS. 


IF  118. 


The  number  of  days  in  each  month  may  easily  be  remembered  from 
the  following  lines : 

Thirty  days  hath  September, 
April,  June,  and  November  ; 
All  the  rest  have  thirty-one, 
Save  February,  which  alone 
.  Hath  twenty-eight ;  and  one  day  more 
We  add  to  it,  one  year  in  four. 

EXAMPLES  FOR  PRACTICE. 


1.  How  many  seconds  from 
Jan.  1,  1790,  till  March  1, 
1804,  including  the  two  days 
named,  and  making  allowance 
for  leap  years  ? 

3.  How  many  minutes  from 
July  4th,  M.,  to  Sept.  29th, 
6  o'clock,  P.  M.  ? 

5.  At  Boston,  on  the  long- 
.est  days,  the  sun  rises  at  23 
min.  past  4  o'clock,  and  sets 
40  min.  past  7;  how  many 
seconds  in  such  a  day? 

7.  How  many  minutes  from 
the  commencement  of  the  war 
between  America  and  Eng 
land,  April  19th,  1775,  to  the 
settlement  of  a  general  peace, 
which  took  place  Jan.  20th, 
1783? 

NOTE  2. — The  pupil  will  no 
tice  that  the  years  1776  and  1780 
were  leap  years. 


2.    In  446947200  seconds 
how  many  weeks  ? 


4.  On  what  month,  day, 
and  hour,  will  125640  minutes 
past  12  o'clock,  M.,  July  4th, 
expire  ? 

6.  In  55020  seconds,  how 
many  hours  ? 


8.    In    4079520 
how  many  years  ? 


minutes, 


Circular  Measure,  or  Motion. 

IT  118.  Circular  Measure  is  used  in  computing  latitude 
and  longitude  ;  also  in  measuring  the  motions  of  the  earth, 
and  other  planets  round  the  sun. 

Questions.  —  Tf  117.  Of  what  is  time  the  measure?  What  are 
the  denominations  ?  Repeat  the  table.  Why  is  1  day  added  to  Feb.  of 
certain  years?  Why  is  it  not  added  to  every  4th  year?  What  years 
have  365  days,  and  what  366  days  ?  What  is  a  year  of  366  days  sailed  ? 
Name  the  calendar  months,  and  the  number  of  days  in  each. 


1T 119,  120.  COMPOUND  NUMBERS.  147 

The  denominations  are  circles,  signs,  degrees,  minutes,  and 
seconds. 

TABLE. 

60  seconds  (")  make  1  minute,  marked  '. 

60  minutes  1  degree,    '    "       °. 

30  degrees  1  sign,  "      S. 

12  signs,  or  360  degrees,  1  circle. 

1.  Reduce  9s.  13°  25'  to  2.  In  1020300",  how  many 
seconds.  degrees  ? 

3.  In  3  signs,  how  many  4.  In  5400',  how  many 
minutes  ?  signs  ? 


IT  119.    Miscellaneous  Table. 

20  units  make  1  score.     100  Ibs.  of  raisins      make  1  cask. 

5  score  1  hundred.      100  Ibs.  of  fish  1  quintal 

12  units  1  dozen.      100  Ibs.  1  hundred 

12  doz.  =  144  1  gross.        18  inches  1  cubit 

12  gross  =  144  doz.  1  great  gross.       22  inches,  nearly,    1  sacred  cubit 

200   Ibs.   of   beef,  )  ,  ,        ,  1  gallon  of  train  oil  7«|  Ibs. 

pork,  or  lish,     j  1  gallon  of  molasses          11  Ibs. 

196  Ibs.  of  flour  1  barrel.       24  sheets  of  paper  1  quire 

8  bushels  of  salt        1  hogshead.       20  quires  1  ream. 

280  Ibs.  of  salt  at  )  .2  reams  1  bundle 

the  salt  works  >  1  barrel.         5  bundles  1  bale. 

in  N.  Y.  ) 

A  sheet  folded  in  two  leaves,  or  4  pages,  is  called  a  folio 

A  sheet  folded  in  four  leaves,  or  8  pages,  a  quarto,  or  4to, 

A  sheet  folded  in  eight  leaves,  or  16  pages,  an  octavo,  or  8vo. 

A  sheet  folded  in  twelve  leaves,  or  24  pages,        a  duodecimo,  or  12mo. 
A  sheet  folded  in  18  leaves,  or  36  pages,  an  18mo. 

A  sheet  folded  in  24  leaves,  or  48  pages,  a  24mo. 

5  points     make  1  line,  )  used  in  measuring  the  length  of  the  rods  of 
12  lines  1  inch,  }         clock  pendulums. 

4  inches  1  hand,  used  in  measuring  the  hight  of  horses. 

6  feet  1  fathom,  used  in  measuring  depths  at  sea. 


Reduction  of  Fractional  Compound  Numbers. 

IF  12O.  There  are  four  particular  cases  in  the  reduction 
of  fractional  compound  numbers.  1st,  To  reduce  a  fraction 
of  a  higher  denomination  to  one  of  a  lower.  2d,  To  reduce  a 
fraction  of  a  lower  denomination  to  one  of  a  higher.  3d,  To 

Questions.  —  IT  118.    What  are  the  uses  of  circular  measure 
What  are  the  denominations  ?    Repeat  the  table. 


148 


COMPOUND  NUMBERS. 


IT  120. 


reduce  a  fraction  of  a  high  denomination  to  integers  of  lower 
denominations.  4th,  To  reduce  integers  of  lower  denomina 
tions  to  a  fraction  of  a  higher.  We  will  consider  them  in 
their  order. 


I.  To  reduce  a  fraction  of 
a  higher  denomination  to  one 
of  a  lower. 

1.  Reduce  -^^  of  a  pound 
to  the  fraction  of  a  penny. 

SOLUTION.  —  We  must  reduce 
2^-Q  of  a  pound  to  the  fraction 
of  a  shilling  by  multiplying  it  by 
20,  since  20  shillings  make  £\. 
This  done  by  ^f  75,  gives  -fa  of 
a  shilling,  which  multiplied  by  the 
composite  number  12,  (If  75,  note 
1,)  is  reduced  to  the  fraction  j- 
of  a  penny.  Hence, 

RULE. 

Multiply  as  in  the  reduction 
of  whole  , numbers,  according 
to  the  rules  for  the  multiplica 
tion  of  fractions. 


II.  To  reduce  a  fraction  of 
a  lower  denominatio?i  to  one 
of  a  higher. 

2.  Reduce  £  of  a  penny  to 
the  fraction  of  a  pound. 

SOLUTION.  —  We  must  reduce 
|-  of  a  penny  to  the  fraction  of  a 
shilling,  by  dividing  it  by  the 
composite  number  12,  since  12 
pence  make  1  shilling.  This 
done  by  *f[  81,  note  2,  gives  -fa 
of  a  shilling,  which  divided  by  20, 
(multiplying  the  denominator,) 
is  reduced  to  the  fraction  5-5-^  of 
a  pound.  Hence, 

RULE. 

Divide  as  in  the  reduction 
of  whole  numbers,  according 
to  the  rules  for  the  division  of 
fractions. 


EXAMPLES  FOR  PRACTICE. 


3.  Reduce  r^  of  a  pound 
of  gold  to  the  fraction  of  a 
grain. 

5.  Reduce  *rfa-Q  of  a  hogs 
head  of  milk  to  the  fraction 
of  a  pint. 

7.  Reduce  ^  of  a  hogs 
head  of  ale  to  the  fraction  of 
a  barrel. 

9.  Reduce  ^f^g-  of  a  tun 
of  oil  to  the  fraction  of  a  gill. 


&•  of  a  grain 
fraction  of  an 


4.    Reduce 
of  gold  to  *the 
ounce. 

6.  Reduce  £gf  of  a  pint  of 
milk  to  the  fraction  of  a  hogs 
head. 

8.  Reduce  -fc  of  a  barrel 
of  ale  to  the  fraction  of  a 
hogshead. 

10.  Reduce  £££f  f  of  a  gill 
of  oil  to  the  fraction  of  a  tun. 


Questions.  — 1[  12O.  How  many  cases  in  the  reduction  of  fractional 
compound  numbers  ?  Give  the  first.  How  are  integers  reduced  from 
a  higher  denomination  to  a  lower  ?  —  from  a  lower  denomination  to  a 
higher  ?  How  are  fractions  reduced  from  a  higher  denomination  to  a 
lower?  Give  the  example  and  its  explanation.  Rule.  How  are  they 
reduced  from  a  lower  denomination  to  a  higher  ?  Give  Ex.  2  and  the 
solution.  Rule. 


If  121. 


COMPOUND  NUMBERS. 


149 


11.  Reduce  TTnT7TTJ.^TnrTTTn7 
of  a  square  m.le  to  the  frac 
tion  of  a  square  inch. 

13.  Reduce  T&T  of  a 
bushel  to  the  fraction  of  a  pint. 

15.  Reduce  -nrijW  of  a 
week  to  the  fraction  of  an 
hour. 

17.  A  cucumber  grew  to 
the  length  of  ^^j  of  a  mile  ; 
what  part  is  that  of  a  foot  ? 

19.  Reduce  $  of  £  of  a 
pound  to  the  fraction  of  a 
shilling. 

21.  Reduce  J  of  ^  of  3 
pounds  to  the  fraction  of  a 
penny  ? 


IT  121.  III.  To  reduce  a 
fraction  of  a  high  denomina 
tion  to  integers  of  lower  de 
nominations. 

1 .  How  many  shillings  and 
pence  in  |  of  a  pound  ? 

SOLUTION. — Multiplying  §-  of 
a.  pound  by  20,  it  is  reduced  to  the 
fraction  of  a  shilling,  &£-.  But 
as  -^P-  of  a  shilling  is  an  improper 
fraction,  (*J[65,)  it  contains  sever 
al  shillings.  The  whole  shillings 
we  find,  dividing  the  numerator 
by  the  denominator,  to  be  13,  and 
a  fraction,  £,  of  a  shilling  remains, 
and  this  reduced  to  the  fraction  of 
a  penny,  is  4/-  of  a  penny  =  4d. 
Hence  13s.  4d.  is  the  Ans. 


13* 


12.   Reduce 
of  a  square  inch  to  the  frac 
tion  of  a  square  mile. 

14.  Reduce  -fa  of  a  pint  to 
the  fraction  of  a  bushel. 

16.  Reduce  ££  of  an  hour 
to  the  fraction  of  a  week. 

18.  A  cucumber  grew  to 
the  length  of  1  foot  4  inches 
=  ||-  =  f  of  a  foot ;  what  part 
is  that  of  a  mile  ? 

20.  f  °-  of  a  shilling  is  f  of 
what  fraction  of  a  pound  ? 

22.  !££.  of  a  penny  is  J  of 
what  fraction  of  3  pounds  ? 
Jj8!2-  of  a  penny  is  ^T  of  what 
part  of  3  pounds  ?  Jj9£  of  a 
penny  is  J  of  -f^  of  how  many 
pounds  ? 

IV.  To  reduce  integers  of 
lower  denominations  to  a  frac 
tion  of  a  higher  denomination. 

2.  What  part  of  a  pound  is 
13s.  4d.  ? 

SOLUTION.  —  In  a  whole  pound 
there  are  240  pence,  and  we  wish 
to  find  what  part  of  this  number 
of  pence  is  contained  in  13s.  4d. 
13s.  4d.  reduced  to  pence,  is  160 
pence.  Hence  13s.  4d.  is  160 
out  of  240  pence  contained  in  a 
whole  pound,  or  ^{y  =  §  of  a 
pound,  Ans. 

NOTE  1 . — The  numerator  and  denom 
inator  of  a  fraction  must  be  of  the  same 
denomination,  since  the  former  is  a 
dividend,  and  the  latter  a  divisor,  both 
of  which  must  be  of  one  denomination, 
(IT  33.)  Hence,  if  there  is  a  frac 
tional  part  to  the  integer  of  the  lowest 
denomination,  for  example,  were  it 
required  to  reduce  4d.  3§qrs.  to  the 
fraction  of  a  shilling,  we  should  have 
to  reduce  Is.  to  Sdsof  a  farthing  fora 


150 


COMPOUND  NUMBERS. 


1T121. 


denominator,  and  4d.  SHqrs.  to  3ds  of 
a  farthing  for  a  numerator.  The  for 
mer  will  then  show  the  number  of  3ds 
of  a  farthing  in  Is.,  the  latter  how 
many  of  them  are  contained  in  4d. 


Herce, 


Hence, 


RULF. 


RULE. 


Reduce  the  given  sum  to 
the  lowest  denomination  con 
tained  in  it  for  a  numerator, 
and  a  unit  of  the  required 
higher  denomination  to  the 
same  denomination  for  the 
denominator. 


Reduce  the  gwen  fraction 
to  the  next  IOWF  t  denomina 
tion  ,  and,  if  it  i-  then  an  im- 
proj.er  fraction,  reduce  it  to  a 
whoie  or  mixed  number, — 
the  integer  is  the  number  of 
this  denomination.  If  a  frac 
tion  remains,  reduce  it,  as 
before,  to  the  next  lower  de 
nomination.  So  proceed,  if  a 
fraction  continues  to  remain, 
to  the  Icwest  denomination. 

EXAMPLES  FOR  PRACTICE. 

3.    .Reduce  f  of  1   day  to         4.    Reduce  14h.  24min.  to 
hours  and  minutes.  the  fraction  of  a  day. 

OPERATION. 

Numer.  3 


OPERATION. 

14A.  24  min.         1  da. 


24 


60 


Dfnom.  5 )  72  ( 14A.  24™.  Am.      864  Numer. 
5 


22 
20 

2 

60 


24 

24 

60 

1440  Denom. 


120 
10 


20 
20 


Questions.  — T[  121.  What  is  case  III.?  Give  the  solution;  the 
lie.  Case  IV. ;  the  solution ;  the  rule.  Why  must  the  numerator  and 
enominator  of  a  fraction  be  of  the  same  denomination  ?  What  follows, 
Hen,  in  case  of  a  fractional  part  ? 


IT  122. 


COMPOUND  NUMBERS. 


161 


5.  What  is  the  value  of  f 
of  a  pound,  Troy  ? 

7.  What  is  the  value  of  f 
of  a  pound,  avoirdupois  ? 

9.  Reduce  f  of  a  mile  to 
its  proper  quantity. 

11.  |  of  a  week  is  how 
many  days,  hours,  and  min 
utes  ? 

13.  Eeduce  -/g-  of  an  acre 
to  its  proper  quantity. 

15.  What  is  the  value  of 
T9^  of  a  yard  ? 


6.  Reduce  7  oz.  4  pwt.  to 
the  fraction  of  a  pound,  Troy. 

8.  Reduce  8  oz.  14f  dr.  to 
the  fraction  of  a  pound,  avoir 
dupois. 

10.  Reduce  4  fur.  125yds. 
2  ft.  1  in.  2}  bar.  to  the  frac 
tion  of  a  mile. 

12.  5  d.  14  h.  24  m.  is 
what  fraction  of  a  week  ? 


14.  Reduce  1  rood  30  poles 
to  the  fraction  of  an  acre. 

16.  Reduce  2  ft.  8  in.  l|b. 
to  the  fraction  of  a  yard. 

NOTE.  —  Let  the  pupil  be  required  to  reverse  and  prove  the  follow 
ing  examples : 

17.  Reduce  3  roods  17£  poles  to  the  fraction  of  an  acre. 

18.  A  man  bought  27  gal.  3  qts.  1  pt.  of  molasses ;  what 
part  is  that  of  a  hogshead  ? 

19.  A  man  purchased  T5^  of  7  cwt.  of  sugar ;  how  much 
sugar  did  he  purchase  ? 

20.  13  h.  42  m.  51f  s.  is  what  part  or  fraction  of  a  day  ? 


Reduction  of  Decimal 

IT  122.  I.  To  reduce  the 
decimal  of  a  higher  denomina 
tion  to  integers  of  lower  de 
nominations. 

1.    Reduce    '375£   to  inte 
gers  of  lower  denominations. 
OPERATION. 
'375£. 
20 


6<000<2.  Ans.  7s.  6d. 

SOLUTION.  —  Multiplying  '375 
of  a  pound  by  20,  it  is  reduced  to 


Compound  Numbers. 

II.  To  reduce  integers  of 
lower  denominations  to  a  deci 
mal  of  a  higher  denomination. 

2.  Reduce  7s.  6d.  to  the 
decimal  of  a  pound. 


OPERATION. 
12    6'0 


20 


7<500 


'375  of  a  pound,  Ans 


SOLUTION.  —  We  divide  6d.  by 
12,  to  reduce  it  to  shillings,  but 


162 


COMPOUND  NUMBERS. 


1T122. 


7'500s.,  observing  the  ordinary 
rule  for  pointing-  of  decimals  in 
the  product,  that  is,  7  shillings, 
and  '500  of  a  shilling.  This 
decimal  multiplied  by  12  becomes 
6'OOOd.,  that  is,  6d.  and  no  deci 
mal.  Hence  '375  of  a  pound  = 
7s.  6d.  Ans. 


Hence, 


RULE. 


Multiply  the  given  decimal 
by  the  number  which  will  re 
duce  it  to  the  next  lower  de 
nomination,  pointing  off  deci 
mals  according  to  the  ordinary 
rule ;  reduce  this  decimal  to 
the  next  lower  denomination, 
pointing  off  as  before.  So 
continue  to  do  through  all  the 
denominations ;  the  several 
integers  will  be  those  re 
quired. 


as  it  will  not  make  a  whole  shil 
ling,  we  annex  a  cipher  to  reduce 
it  to  lOths,  (191;)  then  12  in 
60  tenths,  5  tenths  of  a  shilling ; 
annexing  this  to  7  shillings,  we 
have  75  tenths  of  a  shilling  to  re 
duce  to  the  decimal  of  a  pound, 
and  dividing  by  20,  annexing 
ciphers  to  reduce  the  remainder 
to  hundredths  and  thousandths, 
we  have  '375  of  a  pound. 

Hence, 

RULE. 

Divide  the  lowest  denomi 
nation,  annexing  ciphers  as 
may  be  necessary,  by  the 
number  which  will  reduce  it 
to  the  next  higher,  and  annex 
ing  the  quotient  to  the  num 
ber  of  this  higher  denomina 
tion,  divide  as  before.  So 
continue  to  do  till  the  whole 
is  brought  to  the  required 
decimal. 


EXAMPLES  FOR  PRACTICE. 


3.  Reduce  '213  of  a  "long 
ton  "  to  integers  of  lower  de 
nominations. 

5.  Reduce  '6  of  a  Ib.  of 
emetic  tartar  to  integers  of 
lower  denominations. 

7.  In  '76754  of  a  square 
mile,  how  many  integers  of 
lower  denominations  ? 

9.  Reduce  '3958  of  a  bar 
rel  of  wine  to  integers  of  low 
er  denominations. 


4.  Reduce  4  cwt.  1  qr.  1  Ib. 
1  oz.  14'72  drs.  to  the  deci 
mal  of  a  "  long  ton." 

6.  Reduce  7  g.  1  $.  IB. 
16  grs.  to  the  decimal  fraction 
of  1  Ib. 

8.  What  decimal  of  a 
square  mile  is  491  acres  36 
square  rods  26  square  feet 
19'584  square  inches  ? 

10.  Reduce  12  gal.  1  qt.  1 
pt.  2'9664  gills  of  wine  to  the 
decimal  of  a  barrel. 


Questions.  —  Tf  122. 

mal  compound  numbers? 
solution.  Give  the  rule, 
ample  ?  —  the  rule  ? 


How  many  cases  in  the  reduction  of  deci- 
What  is  case  I.  ?  Give  the  example  and  its 
What  is  case  II.  ?  —  the  solution  of  the  ex 


1T  123.  COMPOUND  NUMBERS.  153 

11.    How  many  integers  of  12.    What  decimal  of  a  cord 

lower  denominations  is  '73  of  is  5C.  ft.  13  cu.  ft.  760'32  cu. 

a  cord  ?  inches  ? 

13.    In  '648  of  a  quarter  of  14.    In  5  bu.   5  qts.   1'776 

wheat,  how  many  integers  of  pts.  of  wheat,  what   fraction 

a  less  denomination  ?  of  a  quarter  ? 

15.    Eeduce  '125  Ibs.  Troy  16.    Reduce  1  oz.   10  pwt. 

to  integers  of  lower  denomi-  to  the  fraction  of  a  pound, 
nations. 

17.    What  is  the  value  of  18.    Eeduce  38  gals.  3'52 

'72  hhd.  of  beer?  qts.  of  beer,  to  the  decimal  of 

ahhd. 

19.    What  is  the  value  of  20.    Reduce  1  qr.  2  na.  to 

'375  of  a  yard  ?  the  decimal  of  a  yard. 

21.    What  is  the  value  of  22.    Reduce  17  h.  6m.  43£ 

'713  of  a  day?  sec.  to  the  decimal  of  a  day. 

Let  the  pupil  be  required  to  reverse  and  prove  the  following  ex 
amples  : 

23.  Reduce  4  poles  to  the  decimal  of  an  acre. 

24.  What  is  the  value  of  '7  of  a  Ib.  of  silver  ? 

25.  Reduce  18  hours   15  m.  50'4  sec.  to  the  decimal  of  a 
day. 

26.  Reduce  11  mi.  6  fur.  2  rods  3  yds.  2  ft.  to  the  decimal 
of  a  degree  on  the  equatorial  circumference  of  the  earth. 


IT  193.     Review   of   Reduction   of  Compound 
Numbers. 

Questions.  —  What  are  compound  numbers?  What  is  meant  by 
the  word  denomination?  What  is  reduction  of  compound  numbers? 
What  are  the  kinds,  and  how  performed?  Changing  ells  Eng.  to  yards 
is  reduction  —  what  kind  ?  What  is  the  use  and  what  the  denomina 
tions  of  Troy  weight  ?  Avoirdupois  weight  ?  Which  is  larger,  1  oz. 
Troy,  or  1  oz.  Avoirdupois? —  1  Ib.  Troy  or  1  Ib.  Avoirdupois?  What 
distinction  do  you  make  between  the  long  and  the  short  ton,  and  where 
are  the  two  used?  What  distinctions  do  you  make  between  linear, 
square,  and  cubic  measure?  What  are  the  denominations  in  linear 
measure  ?  —  in  square  measure  ?  —  in  cubic  measure  ?  How  dc  you 
multiply  by  £  ?  When  the  divisor  contains  a  fraction,  how  do  you  pro 
ceed?  How  is  the  superficial  contents  of  a  square  figure  found?  How 
is  the  solid  contents  of  any  body  found  in  cubic  measure  ?  How  many 
solid  or  cubic  feet  of  wood  make  a  cord?  What  is  understood  by  a  cord 
foot?  ilow  many  such  feet  make  a  cord?  How  many  rods  in  length 
is  Gunter's  chain?  Of  how  many  links  does  it  consist?  How  many 
links  make  a  rod  ?  How  many  rods  in  a  mile  ?  How  many  square 
rods  in  an  acre  ? 


154  COMPOUND  NUMBERS.  1F  123. 

How  many  pounds  make  1  cwt.  ?  For  what  is  circular  measure  used  ? 
Into  (iow  many  parts  is  a  small  circle  divided  ?  —  a  large  circle  ?  —  called 
wha. 

EXERCISES. 

1.  In  £46  4s.  sterling,  how  many  dollars  ?    (Consult  IF  104, 
note  3.)  Ans.  $223'608. 

2.  How  many  rings,  each  weighing  5  pwt.  7  grs.,  may  be 
made  of  3  Ib.  5  oz.  16  pwt.  2  grs.  of  gold  ?  ^i?w.  158. 

3.  Suppose  West  Boston  bridge  to  be  212  rods  in  length, 
how  many  times  will  a  chaise  wheel,  18  feet  6  inches  in  cir 
cumference,  t.urn  round  in  passing  over  it  ? 

Ans.  189/T  times. 

4.  In  10  Ib.  of  silver,  how  many  spoons,  each  weighing  5 
oz.  10  pwt.  ?  .  Ans.  21T^  spoons. 

5.  How  many  sbingles,  each  covering  a  space  of  4  inches 
one  way,  and  6  inches  the  other,  would  it  take  to  cover  1 
square  foot  ?     How  many  to  cover  a  roof  40  feet  long,  and 
24  feet  wide?     (See  IT  48.) 

Ans.  to  the  last,  5760  shingles. 

6.  How  many  cords  of  wood  in  a  pile  26  feet  long,  4  feet 
wide,  and  6  feet  high  ?  Ans.  4  cords,  and  7  cord  feet. 

7.  There  is  a  room  18  feet  long,  16  feet  wide,  and  8  feet 
high;  how  many  rolls  of  paper,  2  feet  wide  and  11  yards 
long,  will  it  take  to  cover  the  walls  ?  Ans.  8^. 

8.  How  many  cord  feet  in  a  load  of  wood  6£  feet  long, 

2  feet  wide,  and  5  feet  high  ?  Ans.  4yJF  cord  feet. 

9.  If  a  ship  sail  7  miles  an  hour,  how  far  will  she  sail  in 

3  w.  4  d.  16  h.  ? 

10.  A  merchant  sold  12  hhds.  of  brandy,  at  $2'75  a  gal 
lon  ;   what  did  he  receive  for  each  hogshead,  and  to  how 
much  did  the  whole  amount  ? 

11.  A  goldsmith  sold  a  tankard  for  £10  8s.  at  the  rate  of 
5s.  4d.  per  ounce  ?  how  much  did  it  weigh  ?  Ans.  3  Ibs.  3  oz. 

12.  An  ingot  of  gold  weighs  2  Ib.  8  oz.  16  pwt. ;  how 
much  is  it  worth  at  3d.  per  pwt.  ? 

13.  If  a  cow  give,  on  an  average,  9  qts.  of  milk  each  day, 
how  much  will  she  give  in  a  year,  or  365  days  ? 

Ans.    15  hhd.  11  gal.  1  qt. 

14.  Reduce  14445  ells  Flemish  to  ells  English. 

15.  There  is  a  house,  the  roof  of  which  is  44J   feet  in 
length,  and  20  feet  in  width,  on  each  of  the  two  sides;  if  3 
shingles  in  width  cover  one   foot  in  length,  how  many  shin 
gles  -vill  it  take  to  lay  one  course  ?     If  3  courses  make  one 


1T  123.  COMPOUND  NUMBERS.  155 

foot,  how  many  courses  will  there  be  on  one  side  of  the  roof? 

How  many  shingles  will  it  take  to  cover  one  side?    to 

cover  both  sides  ?  Ans.  16020  shingles. 

16.  How  many  steps,  of  30  inches  each,  must  a  man  take 
in  traveling  54J  miles  ? 

17.  How  many  seconds  of  time  would  a  person  redeem  in 
40  years,  by  rising  each  morning  J  hour  earlier  than  he  now 
does  ? 

18.  If  a  man  lay  up  4  shillings  each   day,  Sundays  ex- 
cepted,  how  many  dollars  would  he  lay  up  in  45  years  ? 

19.  If  9  candles  are   made   from  1  pound  of  tallow,  how 
many  dozen  can  be  made  from  24  pounds  ? 

20.  If  one   pound  of  wool  make  60  knots  of  yarn,  how 
many  skeins,  of  ten  knots  each,  may  be  spun  from  4  pounds 
6  ounces  of  wool  ?  Ans.  26J  skeins. 

21.  How  many  hours  from  the  commencement  of  the  com 
mon  Christian  era  till  Dec.  10,  1847,  12  o'clock,  noon,  allow 
ance  being  made  for  leap  years  ?     How  many  weeks  ? 

*  A       (  16189932  hours. 
^^  { 98888ft  week.. 

22.  What  part  of  a  pwt.  is  y/^  of  a  pound  Troy  ? 

Ans.  %  pwt. 

23.  What  fraction  of  a  pound  is  |  of  a  farthing  ? 

Ans.  XT^. 

24.  What  fraction  of  an  ell  English  is  4  qrs.*l|  na.  ? 

Ans.  I  E.  E. 

25.  What  fraction  of  a  yard  is  2  qrs.  2|  na.  ? 

Ans.   |  yd. 

26.  What  fraction  of  a  day  is  16  h.  36  m.  55-^  s.  ? 

Ans.  -f%  d. 

27.  What  fraction  of  a  mile  is  6  fur.  26  r.  11  ft.  ? 

Ans.  %  mi. 

28.  What  is  the  value  of  -&  of  a  "  long  ton  ? " 

Ans.  4  cwt.  2  qrs.  12  Ib,  14  oz.  12^  drs. 

29.  What  decimal  of  a  day  is  55  m.  37  sec.  ? 

Ans.  '03862 +d. 

30.  What  decimal  of  a  pound  Troy  is  10  oz.  13  pwt.  9  gr. ? 

Ans.  '8890625. 

31.  What  is  the  value  of  '397  of  a  yard  ? 

Ans.  1  qr.  2  na.  -f-  Ib. 


*  Consult  IT  117,  Note  1. 


156  COMPOUND  NUMBERS.  ^  124 


Addition  of  Compound  Numbers. 

5T  IS4.  1.  A  boy  bought  a  knife  for  1  shilling  9  pence, 
and  a  comb  for  1  shilling  6  pence ;  how  much  did  he  give  for 
both  ?  Am.  3s.  3d. 

2.  A  grocer  sold  at  one  time  2  qts.  of  molasses,  at  another 
time  3  qts.,  at  another  1  qt.,  at  another  3  qts.,  and  at  another 
2  qts. ;  how  many  gallons  did  he  sell  ? 

SOLUTION.  —3  qts.  -f-2  qts.  -{-  1  qt.  ~\-  3  qts.  -f-  2  qts.  =  11  qts., 
and  11  qts.  =  2  gal.  3  qts.  Ans.  2  gal.  3  qts. 

3.  A  boy  had  30  rods  to  walk ;  he  walked  the  first  10  rods 
in  30  seconds,  the  next  10  rods  in  45  sec.,  and  the  last  10 
rods  in  20  sec. ;  how  many  minutes  was  he  in  walking  the 
30  rods  ?  Ans.  1  min.  35  sec. 

4.  What  is  the  amount  of  1  yd.  2  ft.  6  in  -f  2  yds.  1  ft.  8 
m.  ?  A?is.  4  yds.  1  ft.  2  in. 

5.  A  man  has  two  bottles  which  he  wishes  to  fill  with 
wine;  one  will  contain  2  gal.  3  qts.  1  pt.,  and  the  other  3 
qts. ;  how  much  wine  can  he  put  in  them  ? 

Ans.  3  gal.  2  qts.  1  pt. 

The  uniting  together  in  one  sum  of  several  compound  numbers  is 
called  Compound  Addition. 

6.  A  man  bought  a  horse  for  £15  14s.  6d.,  a  pair  of  oxen 
for  £20  2s.  8d.,  and  a  cow  for  £5  6s.  4d. ;  what  did  he  pay 
for  all  ? 

SOLUTION.  —  As  the  numbers  are  large,  we  write  them  down, 
placing  those  of  the  same  denomination  under  each  other,  and,  begin 
ning  with  those  of  the  least  value,  add  up  each  kind  separately ; 
thus  :— 

Then,  adding  up  the  pence,  we  find  the 

IPERATION.         amount  to  be  18,  which  we  divide  by  12  to  re- 

~      -M      «       ^uce  to   shiUmgs ;    t^ie  remainder,   which  is 

pence,  we  write  under  the  column  of  pence, 

20        2     8       and  add  the  1  shilling  to  the  column  of  shil- 

5        &    4       lings.     Adding  up  the   shillings,  we   reduce 

them  to  pounds  ;  setting  the  remainder,  3  shil- 

Ans.    41^  lings,  under  the  column  of  shillings,  we  carry 

the  one  pound  to  the  column  of  pounds,  which 

we  add  up,  setting  down  the  whole  amount,  as  we  do  the  amount  of 
the  last  column  in  simple  addition. 

NOTE.  — The  operations  in  compound  numbers  differ  from  those  of 
simple  numbers  but  in  one  particular.  In  simple  numbers  we  are 


1T  124.  COMPOUND  NUMBERS.  157 

governed  by  the  law  that  figures  increase  in  a  tenfold  proportion  from 
right  to  left,  (*f[  9.)  Compound  numbers  having  no  regular  system 
of  units,  we  are  governed  by  the  relations  between  the  different  de 
nominations  in  which  the  several  quantities  are  expressed. 

The  above  process  is  sufficient  to  establish  the  following 

RULE 

• 

For  Addition  of  Compound  Numbers. 

I.  Write  the  numbers  so  that  those  of  the  same  denomina 
tion  may  stand  under  each  other. 

II.  Add  together  the  numbers  in  the  column  of  the  lowest 
denomination,  and  carry  for  that  number  which  it  takes  of 
the   same  to   make   one   of  the    next   higher  denomination. 
Proceed  in  this  manner  with  all  the  denominations,  till  you 
come  to  the  last,  whose  amount  is  written  as  in  simple  num 
bers. 

PROOF.  —  The  same  as  in  addition  of  simple  numbers. 
EXAMPLES    FOR     PRACTICE. 

7.  8. 

Y.      w.      d.      h.  ra.  s.  T.  cwt.  qr.     Ib.  oz.  dr. 

57      7     6     23  55  11  14  11  1     16  5  10 

84      8     0     16  42  18  25  0  2     11  9  15 

32    24    5       5  18  5  7  18  0    25  11  9 


9.  Bought  a  silver  tankard,  weighing  2  Ib.  3  oz.,  a  silver 
cup,  weighing  3  oz.  10  pwt.,  and  a  silver  thimble,  weighing  2 
pwt.  13  grs. ;  what  was  the  weight  of  the  whole  ? 

Am.  2  Ib.  6  oz.  12  pwt.  13  grs. 

10.  A  ship   landed  at  N.  York  the   following   invoice  of 
English   goods,  viz.,  78  tons  3  cwt.  2  qrs.  26  Ibs.  of  cotton 
goods,   135  tons  15  cwt.  1  qr.  9  Ibs.  of  iron,  ^t)  tons  12  cwt. 
2  qrs.  20  Ibs.  of  woollen  goods,  225  tons  9  cwt.  17  Ibs.  of  coal, 
and    106  tons  1  qr.  of  earthen   ware ;  what  was  the  whole 
amount  ?  Ans.  636  tons  1  cwt.  16  Ibs. 

Questions.  —  If  12*.  What  is  compound  addition  ?  How  are  lower 
denominations  reduced  to  higher  ?  (^  105.}  Repeat  the  rule  for  com 
pound  addition.  How  do  you  carry  1'rom  farthings  to  pence?-  -from 
pence  to  shillings?  —  from  shillings  to  pounds?  How  do  operations  in 
compound  numbers  differ  from  those  in  simple  numbers?  How  lo 
you  carry  through  the  several  denominations  of  avoirdupois  weight? 
—  long  measure  ?  &c..  &c. 
14 


158  COMPOUND    NUMBERS.  1T  124 

NOTE, — It  will  be  recollected,  (^f  106,)  that  what  is  called  tho 
"  long  ton  "  is  used  in  invoices  of  English  goods,  and  of  coal  from 
Pennsylvania. 

11.  A  boat  took  in  freight  as  follows  :  at  one  place,  9576 
Ibs.  of  butter;  at  another,  11  tons  of  pork;  at  a  third,  7  T. 

18  cwt.  27  Ibs.  of  coal ;  what  was  the  entire  freight  in  "  short 
tons  ?  "  Ans.  «4  tons  1299  Ibs. 

12.  A  merchant  bought  3  pieces  of  linen,  measuring  as 
follows :  41  E.  Fl.   1  qr.  2  na.,  18  E.  Fl.  2  qr.  3  na.,  57  E. 
Fl.  1  na. ;  how  many  Flemish  ells  in  the  whole  ? 

Ans.   117E.  Fl.  1  qr.  2  na. 

13.  A  draper  bought  3  pieces  of  English  broadcloth  meas 
uring  as  follows,  viz.,  75  E.  E.  4  qr.  2  na.,  31  E.  E.  1  qr., 
28  E.  E.  1  na. ;  how  many  English  ells  in  the  whole  ? 

Ans.   135  E.  E.  3  na. 

14.  There  are  four  pieces  of  cloth,  which  measure  as  fol 
lows,  viz.,  36  yds.  2  qrs.  1  na.,  18  yds.  1  qr.  2  na.,  46  yds.  3 
qrs.  3  na.,  12  yds.  0  qr.  2  na. ;  how  many  yards  in  the  whole  ? 

Ans.  114  yards. 

15.  A  man  travelled  as  follows,  viz.,  the  1st  day,  35  mi.  7 
fur.  38  rd. ;  2d  day,  4  mi.  2  rd. ;  3d  day,  37  mi.  3  fur.  19  rd.; 
4th  day,  44  mi. ;  what  was  the  length  of  his  journey  ? 

Ans.  121  mi.  3  fur.  19  rd. 

16.  Bought  of  Williams  and  Brother,  London,  1  copy  of 
Shakspeare  for  £7  14s.  6d.,  1  copy  of  Arnold's  Works  for 
£19  10s.  9d.,  1  copy  of  the  Edinburgh  Encyclopedia  for  £27 
6s.,  1  quarto  Bible  for  £8  6d.,  1  copy  of  Johnson's  Works  for 
£15  2s. ;    what  did  the  whole  cost  ?         Ans.  £77  13s.  9d. 

17.  Bought  at  Liverpool  1  bale  of  cotton  goods  for  £9  10s. 
3d.,  1  box  of  jewelry  for  £227  4s.,  1  gross  of  buttons  for  £6 
9s.  8d. ;  what  did  I  pay  for  the  whole  ? 

Ans.  £243  3s.  lid. 

18.  There  are  3  fields,  which  measure  as  follows,  viz.,  17 
A.  3  R.  16  Pf,  28  A.  5  R.  18  P.,  11  A.   25  P. ;    how  much 
land  in  the  three  fields  ?  Ans.  58  A.  1  R.  19  P. 

19.  A  raft  of  hewn  timber  consisted  of  3  cribs ;  the  1st 
crib  contained  29  T.  36  cu.  ft.  1229  cu.  in.  ;  the  2d,  12  T. 

19  cu.  ft.  64  cu.  in. ;  the  3d,  8  T.  11  cu.  ft.  917  cu.  in. ;  how 
much  timber  did  the  raft  contain  ? 

Ans.  50  T.  27  cu.  ft.  482  cu.  in. 

-20.  A  man  removed  79  cu.  yds.  22  cu.  ft.  of  earth  in  dig 
ging  a  cellar,  9  cu.  yds.  26  cu.  ft.  in  digging  a  drain,  and  22 
cu.  yds.  17  cu.  ft.  in  digging  a  cistern  ;  how  much  earth  did 
he  remove?  *An&  112  cu.  yds.  11  cu.  ft 


1T 124.  COMPOUND  NUMBERS.  159 

21.  In  one  pile  of  wood  are  37  cords  119  cu.  ft.  76  cu.  in. ; 
in  another,  9  cords  104  cu.  ft. ;  in  a  3d,  48  cords  7  cu.  ft.  127 
cu.  in. ;  in  a  4th,  61  cords  139  cu.  in. ;  how  much  wood  in 
the  four  piles?  Ans.   156  C.  102  ft.  342  in. 

22.  A  v'ntner  sold  in  one  week  51  hhd.  53  gal.  1  qt.  1  pt. 
of  wine  ;  in  another  week,  27  hhd.  39  gal.  3  qts. ;  and  in 
another  week,  9  hhd.  13  gal.  3  qts. ;  how  much  did  he  sell  in 
the  three  weeks  ?  Ans.  88  hhd.  43  gal.  3  qts.  1  pt. 

23.  A  milk-man  sold  in  one  week,  70  gal.  3  qts.  of  milk ; 
in  another  week,  67  gal.  1  qt. ;  how  many  hogsheads  did  he 
sell  ?  Ans.  2  hhds.  30  gal. 

24.  A  farmer  sowed  36  bush.  2  pks.  5  qts.  1  pt.  of  wheat, 
and  19  bush.  3  pks.  7  qts.  of  barley ;  how  many  bushels  did 
he  sow  ?  Ans.  56  bush.  2  pks.  4  qts.  1  pt. 

25.  A  printer  used  in  one  week  6  bales,  7  reams,  9  quires 
and  9  sheets  of  paper,  and  in  another  week  14  bales,  9  reams, 
19  quires  and  15  sheets ;  how  much  paper  did  he  use  in  the 
two  weeks  ?  Ans.  21  bales  7  reams  9  quires. 

26.  A  ship  sailed  in  one  week  as  follows,  viz.,  on  Mon 
day,  3°  8'  45"  south,  1°  51'  east ;  on  Tuesday,  2°  36'  south, 
2°  T  15"  east ;  on  Wednesday,  4°  52"  south,   1°  east ;  on 
Thursday,  1°  48'  52"  south,  3°  16'  22"  east;  on  Friday,  1° 
19'  south,  48'  29"  east;  and  on  Saturday,  59'  30"  south,  3° 
52'  11"  east;  what  was  her  distance  south  and  east  from  the 
place  of  starting?  A        (  South  13°  52'  59". 

Ans'  I  East  12°  49'  17". 

27.  A  man  plastered  a  church  of  the  following  dimensions, 
viz.,  the  end  walls  contained  116  sq.  yds.  7  sq.  ft.  96  sq.  in. 
each;  the  side  walls,  178  sq.  yds.  138  sq.  in.  each,  and  the 
ceiling  439  sq.  yds.  6  sq.  ft.  78  sq.  in. ;  what  was  the  whole 
amount  of  plastering  in  the  church  ? 

Ans.   1029  sq.  yds.  5  sq.  ft.  114  sq.  in. 

28.  What  is  th,   amount  of  40  weeks  3d.  1  h.  5  m.  +  16 
w.  6  d  4  m.  +  27  w.  5  d.  2  h.  ?       Ans.  85  wk.  3  h.  9  m. 

29.  A  miller  sold  flour  as  follows,  viz.,  4  bar.  176  Ibs.  8 
oz.,  18  bar.  40J  Ibs.,  1  bar.  104  Ibs.  7  oz.,  181|  bar. ;  how 
much  did  he  sell  in  all  ?  Ans.  206  bar.  76  Ibs.  7  oz. 

30.  Five  bags  of  wheat  weighed  as  follows,  viz.,  2J  bush., 
2  bush.  21  Ib.  7  oz.,  1|  bush.  18  Ibs.,  2  bush.  50  Ibs.,  1  bush. 
58|  Ibs.  ;  what  was  their  entire  weight,  calling  60  Ibs.  a 
bushel  ?  Ans.  11  bush.  13  Ibs.  3  oz. 

r 

NOTE. — If  the  four  following-  examples  be  correctly  wrought,  the 
results  will  be  the  same  as  those  here  given! 


160  COMPOUND  NUMBERS.  IT  125,  126. 

31.  What  is  the"  sum  of  35  bar.  27  gal.  3  qts.  +  19  bar. 
5  gal.  1  qt.  +  7  bar.  13  gal.  3  qts.  ? 

Ans.  62  bar.  15  gal.  1  qt. 

32.  What  is  the  sum  of  12  rd.  9  ft.  4  in.  +  15  rd.  7  ft.  8 
in.  -f  6  rd.  4  ft.  5  in.  ?  Ans.  34  rd.  4  ft.  11  in. 

33.  What  is  the  sum  of  the  following  distances  on  the 
equatorial  circumference  of  the  earth,  viz.,  59  deg.  46  mi.  6 
fur.  39  rds.  15  ft.  10  in. ;  216  deg.  39  mi.  7  fur.  39  rds.  4  ft.  7 
in. ;  78  deg.  53  mi.  7  fur.  38  rds.  9  ft.  8  in.  ? 

Ans.  355  deg.  2  mi.  4  fur.  11  rds.  2  ft.  7  in. 

34.  What  is  the  sum  of  2  A.  75  P.  248  sq.  ft.  72  sq.  in. 
-f-  3  A.  120  P.  177  sq.  ft.  85  sq.  in.  -f  15  A.  17  P.  84  sq.  ft. 
80  sq.  in.  ?  Ans.  21  A.  53  P.  238  sq.  ft.  57  sq.  in. 

35.  A  hardware  merchant  sold  several  bills  of  screws,  as 
follows:  25  great  gross  9  gross  7  doz.  11  screws;   15  great 
gross  7  gross  8  doz. ;  40  great  gross  4  doz. ;  what  was  the 
whole  amount  sold  ? 

Ans.  81  gr.  gr.  5  gr.  7  doz.  11  screws. 

Addition  of  Fractional  Compound  Numbers. 

IT  195.     1.  To  £  of  an  hour,  add  £  of  a  minute. 

SOLUTION.  — First  reduce  each  fraction  to  its  proper  quantity  ;  see 
^j  121.  I  of  an  hour  =  52  min.  30  sec.  ;  |  of  a  minute  =  52£  sec. ; 
and  52  min.  30  sec.  -f-  52£  sec.  =  53  min.  22£  sec. 

Ans.  53  min.  22£  sec. 

NOTE.  —  It  may  sometimes  be  more  convenient  to  reduce  the  frac 
tions  to  the  same  denomination,  add  them  together,  and  reduce  their 
fractional  sum  to  its  proper  quantity. 

2.  To  I  of  a  pound,  add  f  of  a  shilling.       Ans.  18s.  3d. 

3.  To  |  of  a  gallon,  add  f  of  a  pint.      Ans.  3  qts.  l-/2-  pts. 

4.  To  I  Ib.  Troy,  add  &  of  an  oz. 

Ans.  6  oz.  11  pwt.  16  gr. 

5.  To  f  of  a  mile,  add  47T3T  rods.    Ans.  239  rds.  4  ft.  6  in. 

6.  To  f  of  20£  yds.,  add  |  of  9£  yds. 

Ans.  15  yds.  1  qr.  2£  na. 


Subtraction  of  Compound  Numbers. 

^T  126.  1.  From  a  piece  of  tape,  containing  9  yds.  3 
qrs.,  sold  4  yds.  1  qr. ;  how  much  remained  ? 

2.  A  woman  having  6  Ibs.  of  butter,  sold  3  Ibs.  10  oz.  •  how 
much  had  she  left  ? 


IT  126.  COMPOUND  NUMBERS.  161 

SOLUTION.  —  Since  there  are  no  oz.  in  the  first  number  or  minu 
end,  we  take  from  6  Ibs.  1  Ib.  =  16  oz.  ;  16  oz.  — 10  oz.  =  6  oz.,  and 
51b.  —  3  Ibs.  =  2  Ibs.  Ans.  2  Ibs.  6  oz. 

3.  How  much  is  1  ft.  —  (less  )  6  in.  ?    1ft.— Sin.?    6ft. 
3  in.  —  1  ft.  6  in.  ?     7  ft.  8  in.  —  4  ft.  2  in.  ?     7  ft.  8  in.  —5 
ft.  10  in.  ? 

4.  How  much  is  4  weeks  3d.  —  3  w.  4  d.  ?     3  w.  1  d.  — 
2w.  5d.? 

Finding  the  difference  between  *two  compound  numbers,  is  called 
Compound  Subtraction. 

5.  From  9  days  15  h.  30  m.,  take  4  d.  9  h.  40  min. 

OPERATION.  SOLUTION.  —  As  the  quantities  are 

d.      h.      m.       large,  it  will  be  more   convenient  to 

Minuend,       9     15     30       write  them  down,  the  less  under  the 

Subtrahend,  4       9     40       greater,  minutes  under  minutes,  hours 

under  hours,  &c.,  since  we  must  sub- 

Ans.    5        5     50       tract  those  of  the  same  denomination 
from  each  other.     We  cannot  take  40 

m.  from  30  m.,  but  we  may,  as  in  simple  numbers,  borrow  from  the 
15  h.  in  the  minuend,  1  hour  =  60  minutes,  which  added  to  30  m.  in 
the  minuend,  makes  90  m.,  and  40  m.  from  90  m.  leaves  50  m., 
which  we  set  down. 

Proceeding  to  the  hours,  having  borrowed  1  from  the  15  h.  in  the 
minuend,  we  must  make  this  number  1  less,  calling  it  14  h.,  and  say, 
9  (hours)  from  14  (hours)  leaves  5,  (hours,)  which  we  set  down. 

Lastly,  proceeding  to  the  days,  4  d.  from  9  d.  leaves  5  d.,  which 
we  set  down,  and  the  work  is  done. 

6.  From  £27,  take  £15  12s.  6d. 

SOLUTION. — We  have  no  pence  from  which   to 
OPERATION.       take  the  6d ^  but  we  mugt  g0  to  the  poun(]S)  an(i  bor- 

^7i,.  s.      d.      row  £l  ==208.,  and  from  the  20s.  borrow  Is.  =  12d. 
15      12     6       Then  6d.  from  12d.  leave  6d. ;  12s.  from  19s.  (which 
1 -I        7     £       remain  of  the  £l)   leave  7s. ;  and  £15  from  £'26 
remaining,  leave  .£11.  Ans.  £11  7s.  6d. 

The  process  in  the  foregoing  examples  may  be  presented 
in  form  of  a 

RULE 

For  Subtracting  Compound  Numbers. 

I.  Write  the  less  quantity  under  the  greater,  placing  simi 
lar  denominations  under  each  other. 

II.  Beginning  with  the  least  denomination,  take  the  lowei 

14* 


162  COMPOUND  NUMBERS.  IF  126, 

number  in  each  from  the  upper,  and  write  the  remainder  un 
derneath. 

III.  If  the  lower  number  of  any  denomination  be  greater 
than  the  upper,  borrow  one  from  the  next  higher  denomination 
of  the  minuend,  reduce  it  to  this  lower  denomination,  subtract 
the  lower  number  therefrom,  and  to  the  remainder  add  the 
upper  number,  remembering  to  call  the  denomination  from 
which  you  borrowed  1  less. 

PROOF.  —  The  same  as  in  simple  subtraction. 

EXAMPLES    FOR    PRACTICE. 

7.  A  London  merchant  sold  goods  to  a  New  York  house 
to  the  amount  of  £136  7s.  6|d.,  and  received  in  payment 
£50  10s.  4|d. ;  how  much  remained  due  ? 

Arts.  £85  17s.  Ifd. 

8.  A  man  bought  a  farm  in  Canada  West  for  £1256  10s., 
and,  in  selling  it,  lost  £87  10s.  6d. ;  how  much  did  he  sell  it 
for?  Ans.  £1168  19s.  6d. 

9.  A  hogshead  of  molasses,  containing  118  gal.,  sprang  a 
leak,  when  it  was  found  only  97  gal.  3  qts.  1  pt.  remained  in 
the  hogshead  ;  how  much  was  the  leakage  ? 

Ans.  20  gal.  0  qt.  1  pt. 

10.  There  was  a  silver  tankard  which  weighed  3  Ib.  4  oz. ; 
the  lid  alone  weighed  5  oz.  7  pwt.  13  grs. ;  how  much  did 
the  tankard  weigh  without  the  lid  ? 

Ans.  2  Ib.  10  oz.  12  pwt.  11  grs. 

11.  From  256  A.  1  R.  10  P.,  take  87  A.  6  P.  10  sq.  yd. 
Ans.  169  A.  1  R.  3  P.  20  sq.  yds.  2  sq.  ft.  36  sq.  in. 

12.  From  15  Ib.  2  oz.  5  pwt.,  take  9  oz.  8  pwt.  10  grs. 

13.  Bought  a  piece  of  black  broadcloth,  containing  36  yds. 
2  qrs. ;  two  pieces  of  blue,  one  containing  10  yds.  3  qrs.  2  na., 
the  other,  18  yds.  3  qrs.  3  na. ;  how  much  more  was  there  of 
the  black  than  of  the  blue  ? 

14.  A  farmer  has  two  mowing  fields ;  one  containing  13 
acres  3  roods,  the  other,  14  acres  3  roods ;  he  has  two  pas 
tures,  also ;  one  containing  26  A.  2  R.  27  P.,  the  other,  45 
A.  2  R.  33  P. ;  how  much  more  has  he  of  pasture  than  of 
mowing  ? 

Questions. — If  126.  What  is  compound  subtraction?  How  do 
you  write  the  qua-ntities? —  the  denominations?  Why  do  yo^.so  write 
them?  When  the  lower  number  of  any  denomination  exceeds  the 
upper,  how  do  you  proceed?  Why?  Repeat  the  rule  for  compound 
subtractioB .  Proof. 


H  127.  COMPOUND  NUMBERS.  163 

NOTE. — If  the  five  following  examples  be  correctly  wrought,  the 
results  will  be  the  same  as  those  here  given. 

15.  From  28  mi.  5  fur.  16  rd.,  take  15  mi.  6  fur.  26  rd.  12 
ft.  Ans.  12  mi.  6  fur.  29  rd.  4  ft.  6  in. 

16.  From  27  P.  16  sq.  ft.  71  sq.  in.,  take  11  P.  110  sq.  ft. 
60  sq.  in.  Ans.  15  P.  178  sq.  ft.  47  sq.  in. 

17.  From  19  P.  55  sq.  ft.  126  sq.  in.,  take  7  P.  92  sq.  ft. 
11  sq.  in.  Ans.  11  P.  236  sq.  ft.  7  sq.  in. 

18.  From  64  A.  2  R.  11  P.  29  sq.  ft.,  take  26  A.  7  R.  34  P. 
132  sq.  ft.    Ans.  36  A.  2  R.  16  P.  169  sq.  ft.  36  sq.  in. 

19.  From  9  rd.  5  yds.  2  ft.  11  in.,  take  10  rd.  0  yd.  1  ft.  2 
in.  ?  Ans.  3  in. 

20.  From  a  pile  of  wood,  containing  21  cords,  was  sold,  at 
one  time,  8  cords  76  cubic  feet;    at  another  time,  5  cords  7 
cord  feet ;  what  was  the  quantity  of  wood  left  ? 

Ans.  6  cords  68  cu.  ft. 

21.  London  is  51°  32',  and  Boston  42°  23'  N.  latitude; 
what  is  the  difference  of  latitude  between  the  two  places  ? 

.      Ans.  9°  9'. 

22.  Boston  is  71°  3  ,  and  the  city  of  Washington  is  77° 
43'  W.  longitude ;  what  is  the  difference  of  longitude  between 
the  two  places  ?  Ans.  6°  40'. 

23.  The  moon  is  8  signs  12°  25'  45"  east  of  the  sun,  and 
Mars  is  11  signs  4°  50'  28"  east  of  the  sun ;  how  far  is  Mars 
east  of  the  moon  ?  Ans.  2  s.  22°  24'  43". 

24.  An  apothecary  had  9fb8i231£13  grs.  of  jalap, 
but  has  used,  in  various  mixtures,  4ft>7§552917  grs. ; 
what  quantity  has  he  left  ?  Ans.  5  ft>  4  5  1  B  16  grs. 

25.  From  124  Ibs.  14  oz.  6  drs.  of  Epsom  salts,  sold  116 
Ibs.  7  oz.  13  drs. ;  what  quantity  remained  unsold  ? 

Ans.  8  Ibs.  6  oz.  9  drs. 

26.  Shipped  to  London  725  qrs.  3  bu.  2  pecks  of  Indian 
corn,  but  unfortunately  218  qrs.  5  bu.  3  pks.  became  damaged; 
what  quantity  remained  uninjured  ? 

Ans.  506  qrs.  5  bu.  3  pks. 

27.  A  ship  loaded  with  615  T.  7  cwt.  of  coal  for  Boston, 
encountering  a  tempest,  a  part  is  thrown  overboard;  there 
were  weighed  out  on  landing,  409  T.  13  cwt.  2  qrs.  27  Ibs. ; 
how  much  was  lost  ?  Ans.  205  T.  13  cwt.  1  qr.  1  Ib. 

IT  127.     Distance  of  TIME  from  one  date  to  another. 

The  distance  of  time  from  one  date  to  another  may  be  found 
by  subtracting  the  first  date  from  the  last,  observing  to  number 
the  months  according  to  their  order. 


164  COMPOUND  NUMBERS.  f  128. 

1.    A  note,  bearing-  date  Dec.  28th,  1846,  was  paid  Jan.  2d, 
1847 ;  how  long  was  it  at  interest  ? 


A.D.  j 


OPERATION. 

1847.     1st  m.     2d  day.  NOTE.  —  In  casting  interest, 

1846.   12      "28      "  an<*  *n  filing  tne  difference  of 

' time  between  dates,  each  month 

Am.     0     0               4  is  reckoned  30  days. 


2.  A  note,  bearing  date  Oct.  20th,  1823,  was  paid  April 
25th,  1825 ;  how  long  was  the  note  at  interest  ? 

3.  What  is  the  difference  of  time  from  Sept.  29th,  1844,  to 
April  2d,  1847  ?  Am.  2  y.  6  m.  3  d. 

4.  A  man  bought  a  farm  April  14th,  1842,  and  was  to  pay 
for  it  Sept.  1st,  1847,  paying  interest  after  Oct.  30th,  1843 ; 
how  much  time  had  he  in  which  to  pay  for  the  farm  ?     How 
much  time  without  interest  ?     For  how  long  a  time  was  he  to 
pay  interest  ?  (  5  yr.  4  m.  17  d. 

Arts.  \\  "     6  "    16  « 
/  3  "  10  "      1  " 


Subtraction  of  Fractional  Compound  Numbers. 

IT  128.     From  f  of  a  week  take  £J  of  a  day. 

SOLUTION.  —  First  reduce  each  fraction  to  its  proper  quantity, 
f  of  a  week  =  2  da.  19  h.  12  m. ;  -f £  of  a  day  =  16  h.  30  m.  ; 
and  2  da.  19  h.  12  m.  — 16  h.  30  m.  =  2  da.  2  h.  42  m. 

Ans.  2  da.  2  h.  42  m. 

NOTE. — .We  may,  if  we  please,  reduce  the  fractions  to  the  same 
denomination,  subtract  them,  and  reduce  their  fractional  difference  to 
its  proper  quantity. 

From  £  of  an  ounce,  take  £  of  a  pwt.    Ans.  11  pwt.  3  grs. 
From  §f  of  a  bushel,  take  -$•%  of  a  peck. 

Ans.  2  pk.  5  qts.  1  pt. 

From  2*5-  of  a  mile,  take  £  of  a  furlong.    Ans.  6  rds.  11  ft. 
From  f  of  19J  gallons,  take  J  of  3|  quarts. 

Ans.  14  gal.  3  qts.  1  pt.  1|  gi. 


Questions.  —  If  127.    How  is  the  distance  of  time  from  one  date 
to  another  foun  I  ?    1  month  is  reckoned  how  many  days  ? 


1T129. 


COMPOUND  NUMBERS. 


165 


Multiplication  and  Division  of  Compound 
Numbers. 

IT  129.     To  multiply  and  divide  by  12  or  less. 

1.    A  man  has  3  pieces  of         2.    If  3  pieces  of  cloth  con- 
cloth,  each  measuring  10yds.     tain  32  yds.  1  qr.,  how  many 
3  qrs. ;  how  many  yds.  in  the     yards  in  one  piece  ? 
whole  ? 

3.    A  man   has   5  bottles, 


each  containing  2  gal.  1  qt.  1 
pt. ;  what  do  they  all  contain  ? 

Hence,  Compound  Multipli 
cation  is,  repeating  a  com 
pound  number  as  many  times 
as  there  are  units  in  the  mul 
tiplier. 

5.  At  1£.  5s.  8|d.  per 
yard,  what  will  6  yards  of 
cloth  cost  ? 


4.  A  man  would  put  11 
gal.  3  qts.  1  pt.  of  vinegar 
into  5  bottles  of  the  same  size, 
what  does  each  contain  ? 

Hence,  Compound  Division 
is,  dividing  a  compound  num 
ber  into  as  many  parts  as  are 
indicated  by  the  divisor. 

6.  If  6  yards  of  cloth  cost 
7£.  14s.  4Jd.,  what  is  the 
price  per  yard  ? 


As  the  numbers  are  large,  we  write  them  down  before  mul 
tiplying  and  dividing. 


OPERATION. 
£.      *.     d.  qrs. 

1     58  3  price  of  I  yard. 
6  number  of  yds. 


OPERATION. 
£.    s.    d.  qrs. 

6)7  14  4  2  cost  of  6  yards. 


Ans.7  14  4  2  cost  of  6  yards. 

SOLUTION.  —  6  times  3qrs.  are 
18qrs.  =4d.  and  2qrs.  over;  we 
write  down  the  2qrs. ;  then,  6 
times  8d.  are  48d.,  and  4  to  car 
ry  makes  52d.  =  4s.  and  4d. 
over ;  we  write  down  the  4d. ; 
again,  6  times  5s.  are  30s.  and 
4  to  carry  makes  34s.  =  \£. 
and  14s.  over;  6  times  l£.  are 
6jC.,  and  1  to  carry  makes  7£., 
which  we  write  down  ;  and  it  is 
plain,  that  the  united  products 
arising  from  the  several  denomi 
nations  is  the  real  product  arising 
from  the  whole  compour  d  number. 


1     583  price  of  1  yard. 

SOLUTION.  —  We  divide  7£. 
by  6,  to  see  how  many  £  each 
yard  will  cost,  and  find  it  to  be 
l£.;  but  6  yards  at  l£.  per 
yard  would  cost  only  6  £ .  Hence , 
the  6  yards  cost  l£.  14s.  4d.  2 
qrs.  more.  Reducing  l£.  to 
shillings,  and  adding  14  shillings, 
we  have  34  shillings,  which,  di 
vided  by  6,  will  give  5  shillings 
more  as  the  price  of  each  yard, 
and  4  shillings  more,  which,  re 
duced  to  pence  and  added  to  4d. 
will  make  52d.,  and  dividing  by 
6,  we  have  8d.  and  4  remainder; 
this  remainder  reduced  to  qrs.  is 
16  and  2  are  18qrs.,  dividing  by  6, 
the  quotient  is  3qrs.  Hence,  each 
yard  will  cost  l£.  5s.  8d.  3qrs. 


166 


COMPOUND  NUMBERS. 


H  129. 


The  processes  in  the  foregoing  examples  may  now  be  pre 
sented  in  form  of  a 


RULE. 


For  multiplying  a  Com 
pound  Number  ivken  the  mul 
tiplier  does  not  exceed  12. 

Multiply  each  denomination 
separately,  beginning  at  the 
least,  as  in  multiplication  of 
simple  numbers,  and  carry  as 
in  addition  of  compound  num 
bers,  setting  down  the  whole 
product  of  the  highest  denomi 
nation. 


For  dividing  a  Compound 
Number  when  the  divisor  does 
not  exceed  12. 

By  short  division,  find  how 
many  times  the  divisor  is  con 
tained  in  the  highest  denomi 
nation,  under  which  write  the 
quotient,  and  if  there  be  a 
remainder  reduce  it  to  the 
next  less  denomination,  add 
ing  thereto  the  number  given, 
if  any,  of  that  denomination, 
and  divide  as  before. 

Proceed  in  this  manner 
through  all  the  denominations, 
and  the  several  quotients  will 
be  the  answer  required. 


EXAMPLES  FOR  PRACTICE. 


7.  What  will  be  the  cost 
of  5  pairs  of  shoes,  at  10s.  6d. 
a  pair  ? 

9.  In  5  barrels  of  wheat, 
each  containing  2  bu.  3  pks. 
6  qts.,  how  many  bushels  ? 

11.  How  many  yards  of 
cloth  will  be  required  for  9 
coats,  allowing  4  yds.  1  qr.  3 
na.  to  each  ? 

13.  In  7  bottles  of  wine, 
each  containing  2  qts.  1  pt. 
3  gills,  how  many  gallons  ? 


8.  At  2£.  12s.  6d.  for  5 
pairs  of  shoes,  what  is  that  a 
pair? 

10.  If  14  bu.  2  pks.  6  qts. 
of  wheat  be  equally  divided 
into  five  barrels,  how  many 
bushels  will  each  contain  ? 

12.  If  9  coats  contain  39 
yds.  3  qrs.  3  na.,  what  does  1 
coat  contain  ? 

14.  If  5  gal.  1  gill  of  wine 
be  divided  equally  into  7  bot 
tles,  how  much  will  each  con 
tain  ? 


Questions.  —  If  129.  What  is  compound  multiplication  ?  Where 
do  you  begin  to  multiply  ?  How  do  you  proceed  when  the  multiplier 
does  not  exceed  12?  How  do  you  carry?  Repeat  the  rule,  when  the 
multiplier  is  12  or  less.  What  is  compound  division?  How  do  you' 
write  the  numbers  when  the  divisor  does  not  exceed  12?  Where  do 
you  begin  the  division  ?  Where  write  the  quotient  ?  If  there  be  a  re 
mainder,  how  do  you  proceed  ?  Repeat  the  rirte. 


H130. 


COMPOUND  NUMBERS. 


167 


15.  What  will  be  the 
weight  of  8  silver  caps,  each 
weighing  5  oz.  12  pwt.  17 
grs.  ? 

17.  How  much  sugar  in  12 
hogsheads,  each  containing  9 
cwt.  3  qrs.  21  Ib.  ? 

19.  How  much  beer  in  9 
casks,  each  containing  1  bar. 
7  gal.  3  qts.  1  pt.  ? 

21.  A  house  has  7  rooms, 
averaging  1  sq.  rod  57  sq.  ft. 
55  sq.  inches ;  what  do  they 
all  contain  ? 

23:  A  boat  on  the  Erie 
canal  averages  21  m.  65  rods 
13  ft.  a  day ;  what  is  that  for 
5  days  ? 

25.  What  quantity  of  land 
in  6  fields,  each  containing 
17  A.  7  sq.  C.  12  sq.  rd.  133 
sq.  1.  ? 

27.  How  much  wood  in  12 
piles,  each  containing  7  C.  5 
c.  ft.  12  cu.  ft.  ? 


16.  If  8  silver  cups  weigh 
3  Ib.  9  oz.  1  pwt.  16  grs.,  what 
is  the  weight  of  each  ? 

18.  If  119  cwt.  Iqr.  of  su 
gar  be  divided  into  12  hogs 
heads,  how  much  will  each 
hogshead  contain  ? 

20.  If  9  equal  casks  of 
beer  contain  10  bar.  34  gal. 
3  qts.  1  pt.,  what  quantity  in 
each? 

22.  If  7  rooms  contain  8 
sq.  rods  129  sq.  ft.  61  sq.  in., 
what  is  the  average  ? 

24.  A  boat  moves  106  m. 
8  rods  15  ft.  6  in.  in  5  days, 
what  is  the  average  of  each 
day? 

26.  If  6  equal  fields  contain 
106  A.  6  sq.  C.  9  sq.  rds.  173 
sq.  1.  ;  how  much  land  in 
each  ? 

28.  In  12  piles  of  wood  are 
92  C.  5  c.  ft.,  how  much  in 
each  ? 


IT  13O.     To  multiply  and 

1.  In  15  loads  of  oats,  each 
measuring  42  bu.  3  pk.  2  qts., 
how  many  bushels  ? 

SOLUTION.  —  Multiplying  the 
quantity  in  1  load  by  3,  we  have 
the  quantity  in  3  loads,  and  mul 
tiplying  the  quantity  in  3  loads  by 
5,  we  have  the  quantity  in  5  times 
3,  or  15  loads. 


divide  by  a  composite  number. 

2.  If  15  loads  of  oats  meas 
ure  642  bu.  0  pk.  6  qts.,  how 
many  bushels  in  each  load  ? 

SOLUTION.  —  Dividing  the 
quantity  in  15  loads  by  5,  we 
have  the  quantity  in  £  of  15,  or 
3  loads,  and  dividing  the  quantity 
in  3  loads  by  3,  we  have  the 
quantity  in  1  load. 


Questions.  —  If  130.  Wh<»^  the  multiplier,  or  divisor,  is  a  com 
posite  number,  how  may  the  operation  be  contracted  in  multiplication? 
—  in  division  ? 


188 


COMPOUND  NUMBERS. 


1T131 


OPERATION. 
bu.  pk.  qts. 

42  3  2  in  I  load. 
3  one  factor. 


128 


1  6  in  3  loads. 
5  the  other  factor. 


OPERATION. 

bu.  pk.  qts. 

/(J£|5)642  0  6  in  15  loads 


3/«c°/£13)128  1  6  in  3  loads. 
Ans.  42  3  2  in  1 


Hence,  W/ferc  the  multiplier  or  divisor  exceeds  12  and  is  a 
composite  number, 

RULE. 

Multiply   by   each   of    the         Divide  by  each  of  the  corn- 

component  parts  of  the  multi-  ponent  parts  of  the   divisor. 

plier.     The  last  product  will  The  last  quotient  will  be  the 

be  the  answer.  answer. 


642  0  6  m  15 


3.  What  will  24  barrels  of 
flour  cost,  at  2£.  12s.  4d.  a 
barrel ? 

5.  How  many  bushels  of 
apples  in  112  barrels,  each 
barrel  containing  2  bu.  1  pk.  ? 

NOTE.  —  8, 7,  and  2,  are  factors 
of  112. 

7.  How  much  molasses  in 
84  hogsheads,  each  hogshead 
containing  112  gal.  2  qts.  1 
pt.  3  gi.  ? 

9.    How  many  bushels  of 
wheat  in  135  bags,  each  con 
taining  2  bu.  3  pecks  ? 
3X9X5=  135. 

11.  Sold  25  pieces  of  cloth, 
each  containing  32  yds.  2  qr. 
1  na. ;  how  much  in  the 
whole  ? 


4.  Bought  24  barrels  of 
flour,  for  62<£.  16s. ;  how 
much  was  that  per  barrel  ? 

6.    In  112  barrels  are  252 

bushels  of  apples  ;  how  many 
bushels  in  1  barrel  ? 


8.  Bought  84  hogsheads 
of  molasses,  containing  9468 
gal.  1  qt.  1  pt. ;  how  much  in 
a  hogshead? 

10.  371bu.  Ipk.  of  wheat 
are  equally  divided  into  135 
bags  ;  how  much  in  each  ? 

12.  Sold  814  yds.  1  na.  of 
cloth  in  25  equal  pieces ;  how 
much  in  each  piece  ? 


IT  131.      To  multiply  and  divide  by  any  number  greatei 
than  12,  which  is  not  a  composite  number. 

1.    How    many   yards    of  2.    Bought   139   pieces  of 

sheeting  in  139  pieces,  eacr  sheeting,  containing  4439  yds', 

piece  containing  31  yds.  3  qrs.  i  qr.  1  na. ;  how  many  yards 

3  na.  ?  in  1  piece  ? 


IT  131. 


COMM)UND  NUMBERS. 


169 


SOLUTION.  —  139  is  not  a  com 
posite  number.  We  may,  how 
ever,  decompose  this  number  thus, 
]39=lOp-f  30-J-9. 

We  may  now  multiply  the  num 
ber  of  yards  in  1  piece  by  10, 
which  will  give  the  number  of 
yards  in  10  pieces,  and  this  prod 
uct  again  by  10,  which  will  give 
the  number  of  yards  in  100  pieces. 

We  may  next  multiply  the 
number  of  yards  in  10  pieces  by 
3,  which  will  give  the  number  of 
yards  in  30  pieces,  and  the  number 
of  yards  in  1  piece  by  9,  which 
will  give  the  number  of  yards  in  9 
pieces,  and  these  three  products, 
added  together,  will  evidently 
give  the  number  of  yards  in  139 
pieces ;  thus : 

yds.    qrs.    na. 

31     3     3  in  I  piece. 
10 


319     1    2  in  10  pieces. 

10 


3193     3     0  in  100  pieces. 
958     0     2  in    30  pieces. 
1     3  in      9  pieces. 


ww 

237 


4439     1     1  in  139  pieces. 

NOTE. — In  multiplying  the 
number  of  yards  in  10  pieces, 
(319yds.  1  qr.  2  na.,)  by  3,  to 
get  the  number  of  yards  in  30 
pieces,  and  in  multiplying  the 
number  of  yards  in  1  piece,  (31 
yds.  3  qrs.  3  na.,)  by  9,  to  get 
the  number  of  yards  in  9  pieces, 
the  multipliers,  3  and  9,  need  not 
be  written  down. 


SOLUTION.  — When  the  divisor 
cannot  be  produced  by  the  multi 
plication  of  small  numbers,  the 
better  way  is  to  divide  after  the 
manner  of  long  division,  setting 
down  the  work  of  dividing  and  re 
ducing  as  follows  : 

yds       qT.  na.  yds.  (jr.  na. 

139)4439  1  1(31  3  3 
417 

269 
139 

130 
4 

521  (3  qrs. 
417 

104 
4 


417 

We  divide  4439  yds.  by  139,  to 
ascertain  the  number  of  yards  in 
each  piece,  which  we  find  to  be 
31,  and  a  remainder  of  130  yds., 
which  will  not  be  a  whole  yard  to 
each  piece  ;  but  reducing  them  to 
quarters,  adding  in  the  1  qr.,  wo 
have  521  qrs.,  which  we  divide  by 
139  to  ascertain  the  number  of 
quarters  in  each  piece,  and  find  it 
to  be  3,  and  a.  remainder  of  104 
qrs.  ;  we  reduce  this  remainder  to 
nails,  adding  in  the  1  nail,  and 
have  417  na.,  which  we  divide  by 
139  for  the  number  of  nails  in 
each  piece,  and  find  it  to  be  3, 
without  a  remainder.  Hence, 
there  are  31  yds.  3  qrs.  3  na.  in 
each  piece,  Ans. 


Questions.  —  ^f  131.    When  the  multiplier  or  divisor  exceeds  12, 
how  is  the  multiplication  performed  ?  —  the  division  ? 
15 


170 


COMPOUND  NUMBERS. 


1T131 


Hence,  When  the  multiplier  or  divisor  exceeds  12,  and  is 
not  a  composite  number, 

RULE. 


Multiply  first  by  10,  and 
this  product  by  10,  which  will 
give  the  product  for  100 ;  and 
if  the  hundreds  in  the  multi 
plier  be  more  than  one,  multi 
ply  the  product  of  100  by  the 
number  of  hundreds  ;  for  the 
tens,  multiply  the  product  of 
10  by  the  number  of  tens  ;  for 
the  units,  multiply  the  multi 
plicand;  and  the  sum  of  these 
several  products  will  be  the 
product  required. 


Divide  after  the  manner  of 
long  division,  setting  down 
the  work  of  dividing  and 
reducing. 


EXAMPLES  FOR   PRACTICE. 


3.  How  many  acres  in  241 
wild  lots,  each  containing  75 
acres  2  roods  25  rods  ? 


5.  How  many  pounds  of 
tea  in  23  chests,  each  contain 
ing  78  Ibs.  9  oz.  ? 

NOTE.  — The  pupil  will  easily 
perceive  the  method  of  operation, 
when  the  multiplier  is  less  than 
100. 

7.  Bought  375  bales  of 
English  goods  at  9£.  11s.  6d. 
per  bale  ;  what  did  the  whole 
cost? 

9.  How  many  bushels  of 
wheat  are  raised  on  125  acres, 
averaging  22  bush.  3  pecks  5 
quarts  to  the  acre  ? 


4.  There  are  surveyed  in 
an  unsettled  district,  18233 
acres  25  rods  of  land,  which 
is  divided  into  241  equal  lots  ; 
how  many  acres  in  each  lot  ? 

6.  If  1806  Ibs.  15  oz.  of  tea 
be  divided  equally  into  23 
chests,  how  much  will  be  in 
each  chest  ? 


8.  Bought  375  bales  of 
English  goods  for  3590£.  12s. 
6d. ;  what  did  each  bale  cost  ? 

10.  A  wealthy  farmer  har 
vested  125  acres  of  wheat, 
which  yielded  2863  bush.  1 
peck  1  qt. ;  what  was  the 
average  per  acre  ? 


IT  132. 


COMPOUND  NUMBERS 


171 


IT  133.    Difference  in  longitude  and  time  between  different 
places. 

Every  circle,  whether  great  or  small,  is  supposed  to  be  di 
vided  into  360  equal  parts,  called  degrees. 

Let  the  accompanying  dia 
gram  represent  the  great  cir 
cle  of  the  earth,  called  the 
equator,  divided,  as  you  he  e 
see,  into  24  equal  parts  of  15 
degrees  each,  (360°  ~-  24  = 
15°.) 

As  the  sun  apparently 
passes  round  the  earth  in  24 
hours,  it  will  pass  through 
one  of  tjiese  divisions,  or  15°, 
in  1  hour  =  60  minutes  of 
time,  and  of  course  it  will 
pass  1°  of  motion  in  fa  of  60  ==  4  minutes  of  time,  and  1'  of 
motion  in  FV  of  4  minutes  (=  240  seconds  -r-  60)  =  4  seconds 
of  time.  Hence  it  follows,  that  the  apparent  motion  of  the 
sun  round  the  earth,  from  east  to  west,  is 

15°  of  motion  in  1  hour  of  time, 
1°  of  motion  in  4   minutes  of  time,  and 
1'  of  motion  in  4  seconds  of  time. 


From  these  premises, 
there  is  a  difference  in  longi 
tude  between  two  places  there 
will  be  a  corresponding  differ 
ence  in  the  hour,  or  time  of 
the  day.  The  difference  in 
longitude  being  15°,  the  dif 
ference  in  time  will  be  1  hour; 
the  place  easterly  having  noon, 
or  any  other  specified  time,  1 
hour  sooner  than  the  place 
westerly. 

Hence,  if  the  difference  in 
longitude,  in  degrees,  and 
minutes,  be  multiplied  by  4, 
the  product  will  be  the  differ 
ence  in  time  in  minutes  and 
seconds,  which  may  be  re 
duced  to  hours 


it  follows  that  when 
there  is  a  difference  in  time 
between  two  places,  there  is  a 
corresponding  difference  in 
their  longitude.  If  the  differ 
ence  in  time  be  1  hour,  the 
difference  in  longitude  will  be 
15°  ;  if  4  minutes,  the  differ 
ence  in  longitude  will  be  1°, 
&c. 

Hence,  if  the  difference  in 
time  (in  minutes  and  seconds) 
between  two  places  be  divided 
by  4,  the  quotient  will  be  the 
difference  in  longitude,  in  de 
grees  and  minutes. 


172  COMPOUND  NUMBERS.  U"  133 

1     What  is  the  difference  2.    What  is  the  difference 

in  time  between  London  and  in  longitude  between  London 

Washington,  the  difference  in  and   Washington,  the    differ- 

longitude  being  77°  ?  ence  in  time  being  5  h.  8m.? 

SOLUTION.  —  The  sun  appar-  SOLUTION.  —  The  sun's  appar 
ently  moves  1°  in  4  minutes  of  ent  motion  is  1°  in  4  minutes  of 
time,  and  it  will  move  77°  in  77  time,  and  we  wish  to  know  how 
times  4  minutes  =  308  minutes,  far  it  will  move  in  308  minutes. 
Or,  we  may  multiply  77  by  4,  It  will  move  as  many  degrees  as 
since  either  factor  may  he  the  4  min.  is  contained  times  in  (can 
multiplicand.  And  308  min.  =  be  subtracted  from)  308  min 
6  h.  8  miii,  Ans.  And  308  ^-  4  =  77°,  Ans. 

3.  When  it  is  12  o'clock  at  4.  When  it  is  12  o'clock  at 
the  most  easterly  extremity  of  the  most  easterly  extremity  of 
the  island  of  Cuba,  what  will  the  island  of  Cuba,  it  is  16 
be  the  hour  at  the  most  wes-  minutes  past  11  o'clock  at  the 
terly  extremity,  the  difference  western  extremity  ;  what  is 
in  longitude  being  11°  ?  the  difference  in  longitude  be 

tween  the  two  points? 

5.  Supposing  a  meteor  should  appear  so  high  that  it  could 
be  seen,  at  once  by  the  inhabitants  of  Boston,  71°  3',  of  Wash 
ington,  77°  43',  and  of  the  Sandwich  Islands,  155°  west  lon 
gitude  ;  if  the  time  be  47  minutes  past  11  o'clock  of  Dec.  31, 
1847,  at  Washington,  what -will  be  the  time  at  Boston,  and  at 
the  Sandwich  Islands  ? 

Am.  At  Boston,  13  min.  40  sec.  A.  M.  (morning)  of  Jan. 
1st,  1848 ;  at  the  Sandwich  Islands,  37  miri.  52  sec.  past  6 
o'clock,  P.  M.,  of  Dec.  31,  1847. 


f  133.    Review  of  Compound  Numbers. 

Questions. — What  distinction  do  you  make  between  simple  and 
compound  numbers?  (^[  102.)  What  is  the  rule  for  adding  compound 
numbers  ?  In  Tf  124,  Ex.  33,  what  difficulty  is  met  with  in  carrying 
from  miles  to  degrees  ?  How  is  it  obviated?  Rule  for  subtracting  coin- 
Questions.  —  "[[  132.  What  circle  is  the  diagram  intended  to  rep 
resent  ?  Into  how  many  divisions  is  it  divided  ?  how  many  degrees  in 
each  division  ;  and  what  does  it  represent?  In  what  time  does  the  sun 
move  1°,  and  why  ?  —  1',  and  why  ?  What  does  M. "signify  ?  A-  M.  ? 
P.  M.?  What  motion  causes  the  apparent  motion  of  the  sun?  When 
the  difference  of  longitude  between  places  is  known,  how  may  the  differ 
ence  in  time  be  calculated  ?  When  it  is  noon  at  any  place,  is  it  before 
noon  or  after  noon  at  places  easterly?  —  at  places  westerly?  Why? 
When,  there  is  a  difference  in  time  between  places,  what  follows? 
When  the  difference  in  time  is  given,  how  may  the  difference  in  longi 
tude  be  found? 


IF  133.  COMPOUND  NUMBERS.  173 

pound  numbers?  —  for  multiplying  when  the  multiplier  does  not  exceed 
12  ?  —  when  it  does  exceed  12,  and  is  a  composite  number  ?  —  w^en  not  a 
composite  number  ?  —  for  dividing  compound  numbers,  when  the  divisor 
does  not  exceed  12?  —  when  it  exceeds  12  and  is  a  composite  number? 
—  when  not  a  composite  number  ?  How  is  the  distance  of  time  from  one 
date  to  another  found  ?  How  many  degrees  does  the  earth  revolve  from 
west  to  east  in  1  hour?  In  what  time  does  it  revolve  1°?  Where  is 
the  time  or  hour  of  the  day  sooner  —  at  the  place  most  easterly  or  most 
westerly  \  The  difference  in  longitude  between  two  places  being  known, 
how  is  the  difference  in  time  calculated  ?  The  difference  in  time  being 
known,  how  is  the  difference  in  longitude  calculated  ? 

EXERCISES. 

1.  A  gentleman  is  possessed  of  l£  dozen  of  silver  spoons, 
each  weighing  3  oz.  5  pwt. ;  2  doz.  of  tea-spoons,  each  weigh 
ing  15  pwt.  14  gr. ;   3  silver  ciins,  each  9  oz.  7  pwt.  ;  2  silver 
tankards,  each  21  oz.  15  pwt. ;    and  6  silver  porringers,  each 

11  oz.  18  pwt. ;  what  is  the  weight  of  the  whole  ? 

Am.  18  Ib.  4  oz.  3  pwt. 

2.  In  35  pieces  of  cloth,  each  measuring  27  yds.  3  qrs., 
how  many  yards  ?  Ans.  971  yds.  1  qr. 

3.  How  much  wine  in  9  casks,  each  containing  45  gal.  3 
qts.  1  pt.  ? 

4.  If  a  horse  travel  a  mile  in  12  min.  16  sec.,  in  what  time 
would  he  travel  176  miles  ?      Am.  Id.  1 1  h.  58  m.  56  sec. 

5.  If  8  horses  consume  889  bu.  2  pks.  6  qts.  of  oats  in  365 
days,  how  much  will  one  horse  consume  in  1  day? 

Ans.  1  pk.  1  qt.  1  pt.  2  gills. 

6.  I  hold  an  obligation  against  George  Brown,  of  London, 
of  735£.  11s.  6d.,  on  which  are  two  endorsements,  viz.,  61£. 
5s.  and  195£.  13s.  lid. ;  what  remains  unpaid? 

Ans.  478£.  12s.  7d. 

7.  Liverpool,  Jan.  1,  1848. 
Isaac  Derwent,  of  Boston,  U.  S. 

Bought  of  Shipley  &  Co. 
10  boy's  hats,  No.  1,  4s.  6d. 

12  do.  2,  5s. 

4        do.  3,  5s.  6d. 

4         do.  9,  10s. 

4         <b,         ~    10,  11s. 

6         do.  11,  12s. 

6  men's  hats,  14s. 

\  trunk  for  packing,  1£.  4s. 


19£.  11s. 
Received  payment,  Shipley  &  Co. 

15*  by  G.  Willkma. 


174  ANALYSIS.  IT  134. 

NOTE.  —  If  the  ,hr3e  following  examples  bewiought  correctly,  the 
answers  will  be  as  here  given. 

8.  A  man  divides  16  bar.  23  gal.  3  qts.  of  oil  into  5  large 
vessels ;  how  much  does  he  put  in  each  ? 

Ans.  3  bar.  11  gal.  0£  qt. 

9.  On  an  acre  of  ground  were  erected  21  buildings,  occu 
pying  on  an  average  3  sq.  rds.  112  ft.  81  in. ;  how  much  re 
mained  unoccupied  ?  Ans.  2  roods  8  rods  86  ft.  63  in. 

10.  If  a  man  build  3  rds.  9  ft.  7  in.  in  length  of  wall  in  1 
day,  how  many  rods  can  he  build  in  15  days  ? 

Ans.  53  rds.  11  ft.  9  in. 

11.  If  a  ship  sail  3°  18'  45"  in  one  day,  how  far  will  it 
nail  in  the  month  of  June  ?  Ans.  99°  22'  30". 

12.  If  a  druggist  sell  1  gross  7  doz.  bottles  of  sarsaparilla 
in  1  week,  how  many  gross  will  he  sell  in  the  months  of 
April,  May  and  June,  at  the  same  rate  ?    Ans.  20  gr.  7  doz. 

13.  If  a  man,  employed  in  counting  money  from  a  heap, 
count  100  silver  dollars  in  a  minute,  and  continue  at  the  work 
10  hours  each  day,  how  many  days  will  it  take  him  to  count 
a  million  ?  Ans.  16§  days. 

14.  At  the  same  rate,  how  many  years,  reckoning  365 
days  to  a  year,  would  it  take  him  to  count  a  billion  ? 

Ans.  45  years  241|  days. 

15.  Were  1000  men  employed  at  this  same  business,  each 
one  counting  at  the  same  rate,  10  hours  each  day,  how  many 
years  would  it  take  them  to  count  a  quadrillion  ? 

Ans.  45662  years  36f  days. 


ANALYSIS. 

5T  134.  In  most  examples  in  arithmetic,  two  things  are 
given  to  find  a  third.  Thus,  in  the  relations  of  the  price  and 
quantity,  the  quantity  and  the  price  of  a  unit  may  be  given  to 
find  the  price  of  the  quantity,  or  the  quantity  and  its  price  to 
find  the  price  of  a  unit,  or  the  price  of  a  unit  and  of  a  quan 
tity  to  find  the  quantity.  The  same  principles  may  readily 
be  applied  to  other  calculations. 

This  method  of  operating  is  called  Analysis.  Analysis, 
therefore,  may  be  defined,  the  solving  of  questions  on  general 
principles. 

We  have  presented  (IT  46)  these  rules  in  connection,  as  ap 
plied  to  whole  numbers,  and  separately  as  applied  to  any 


1T134. 


ANALYSIS. 


175 


quantities  in  HIT  77, 
together  as  applied 
mutually  prove  each 
tional  losses. 

I.  The  price  of 
unity,  and  the  quan 
tity  being  given,  to 
find  the  price  of  the 
quantity, 

RULE. 

Multiply  the  price 
by  the  quantity. 


85,  and  83.  We  will  now  present  them 
to  any  quantities,  with  examples  which 
other,  except  where  there  are  some  frac- 


II.  The  quanti 
ty,  and  the  price  of 
the  quantity  being 
given,  to  find  the 
price  of  unity, 


III.  The  price  of 
unity,  and  the  price 
of  a  quantity  being 
given,  to  find  the 
quantity, 


RULE, 


RULE. 


Divide   the  cost 
by  the  quantity. 


Divide  the  price 
of  the  quantity  by 
the  price  of  unity. 


EXAMPLES    FOR    PRACTICE. 


1.  At  $302'40 
per  tun,  what  will 
Ihhd.  15  gal.  3  qts. 
of  wine  cost  ? 

4.  At$2<215per 
gal.,  what  cost  3J 
qts.? 

7.  At$96'72per 
ton  for  pot-ashes, 
what  will  f  of  a  ton 
cost? 

10.  If  a  yard  of 
cloth  cost  $2'5, 
what  will  '8  of  a 
yard  cost  ? 

13.  If  a  ton  of 
pot-ashes  cost  £27 
10s.,  what  will  14 
cwt.  cost? 

16.  If  a  bushel 
of  wheat  cost 
$1'92,  what  will  1 
pk.  4  qts  cost  ? 


2.  At  $94<50 
for  1  hhd.  15  gal. 
3  qts.  of  wine, 
what  is  that  per 
tun? 

5.  At  Sl'80  for 
3J  qts.  of  wine, 
what  is  that  per 
gal.? 

8.  If  |  of  a  ton 
of  pot-ashes  cost 
$60'45,  what  is  that 
per  ton  ? 

11.  If  '8  of  a 
yard  of  cloth  cost 
$2,  what  is  that  per 
yard  ? 

14.  If  14  cwt.  of 
pot-ashes  cost  £19 
5s.,  what  is  that 
per  ton  ? 

17.  If  1  pk.  4 
qts.  of  wheat  cost 
$'72,  what  is  that 
per  bushe  ? 


3.  At  $302<40 
per  tun,  how  much 
wine  may  be  bought 
for  $94<50. 

6.  At$2'215per 
gal.,  how  much 
wine  may  be  bought 
for  $1<80  ? 

9.  At$96<72per 
ton,  how  much  pot 
ashes  may  be 
bought  for  $60'45  ? 

12.  At  $2<5  per 
yard,  how  much 
cloth  may  be  pur 
chased  for  $2  ? 

15.  At  £27  10s. 
a  ton  for  pot-ashes, 
what  quantity  may 
be  bought  for  £19 
5s.? 

18.  At  $1<92 
per  bushel,  how 
much  wheat  may 
be  bought  for  $'72? 


176 


ANALYSIS. 


IT  134 


19.  If  a  yard  of 
broadcloth  cost  $6, 
what  will  16  yds. 
2  qrs.  3  na.  cost  ? 

22.  If  a  ton  of 
hay  cost  $13,  what 
will  1850  Ibs.  cost? 


2-5.  If  an  eagle 
weigh  11  pwt.  6 
grs.,  what  will  be 
the  weight  of  665 
eagles  ? 


20.  If  16  yds.  2 
qrs.  3  na.  of  broad 
cloth  cost  $ 100'  125, 
what  is  that  per 
yard  ? 

23.  If  1850  Ibs. 
of  hay  cost  $12- 
'025,  what  is  that 
per  ton  ? 

26.  If665eagles 
weigh  31  Ibs.  2  oz. 
1  pwt.  6  grs.,  what 
is  the  weight  of 
each? 


21.  At  $6  per 
yard,  how  much 
broadcloth  may  be 
bouo-ht  for  $100- 
'125? 

24.  At  $13  per 
ton,  how  many  Ibs, 
of  hay  may  be 
bought  for  $12* 
'025  ? 

27.  How  many 
eagles,  each  weigh 
ing  11  pwt.  6  grs., 
may  be  coined  from 
31  Ibs.  2  oz.  1  pwt. 
6  grs.  of  standard 
gold? 

NOTE.  —  After  the  same  manner  let  the  pupil  reverse  the  follow 
ing  examples : 

28.  At  $1'75  a  bushel  for  wheat,  how  many  quarters  car 
be  bought  for  $200  ?     Ans.  11  qrs.  2  bu.  1  pk.  l'1428f  qts. 

29.  What  will  3  qrs.  2  na.  of  broadcloth  cost,  at  $6  pe 
yard  ? 

30.  At  $22' 10  for  the  transportation  of  6500  Ibs.  46  mile* 
what  is  that  per  ton  ?  $6'80. 

31.  Bought  a  silv  r  cup,  weighing  9  oz.  4  pwt.  16  grs.,  fc* 
$1 1'OS ;  what  was  tl  it  per  ounce  ? 

32.  A  lady  purchased  a  gold  ring,  giving  at  the  rate  o\ 
$20  per  ounce  ;  she  paid  for  the  ring  $T25;  how  much  d>d 
it  weigh  ?  Ans.  1  pwt.  6  grs. 

Examples  requiring  several  operations. 

33.  If  6  bushels  of  wheat  cost  $7'50,  what  will  28  bushes 
cost  ? 

NOTE  1.  It  was  customary  formerly  to  perform  all  examples  slcni- 

Questions.  —  ^[  134.  What  do  you  say  of  every  example  in  arith 
metic  ?  What  severally  are  given  and  required  in  the  relations  of  the 
price  and  quantity  ?  Answer  the  questions  in  ^[  46.  (  The  teacher  nil::  ask 
them.)  What  is  given  and  what  required  under  I.  ?  —  the  rule  ?  —  under 
II.  ?  —  the  rule  ?  —  under  III.  ?  —  the  rule  ?  How  were  such  examples  as 
the  33d  formerly  wrought  ?  What  method  is  preferable  ?  What  do  we  do 
by  this  method  ?  Explain  the  solution  of  this  example  ;  —  the  first  solu 
tion  of  example  34  ;  —  the  second  method.  What  is  this  called  ?  "What 
is  an  analytic  solution  ? 


IT  134.  ANALYSIS.  177 

lar  to  the  above  by  a  rule  called  the  Rule  of  Three,  or,  on  account  of 
its  supposed  importance,  the  golden  rule.  But  they  are  more  intelligi 
bly  solved  by  analysis ;  that  is,  from  the  things  given  in  the  question, 
find  the  price  of  unity,  and  having  the  price  of  unity,  find  the  price  of 
the  quantity,  the  price  of  which  is  required.  Thus, 

OPERATION. 


$1*25,  1  bu.          SOLUTION.  —  Dividing  the  price  of  6 

bu.  by  6  will  give  the  price  of  1  ba., 
arid  multiplying  the  price  of  1  bu.  by 
1^uu  28  will  give  the  price  of  28  bu. 


34.  If  T£  of  a  bushel  of  com  cost  ^  of  a  dollar,  what  will 
|§  of  a  bushel  cost  ? 

SOLUTION.  —  We  divide  the  price  of  the  quantity,  ^,  by  the 
quantity,  y^-,  to  get  the  price  of  unity  :  y7-^  -f-  i^1^^^  °^  a  ^°^" 
*ar,  price  of  1  bu.  And  multiplying  the  price  of  1  bu.  by  |-§  of  a 
bushel,  will  give  the  price  of  the  quantity  required  :  y9^  X  £  f  = 
f  §£f  of  a  dollar. 

Or,  to  give  a  more  full  analysis,  if  y^£  of  a  bushel  cost  y7-^  of  a 
dollar,  3*3-  would  cost  TT  as  much  :  TT  of  y7-^  =  T^  ;  and  Tf  = 
1  bu.,  would  cost  13  times  as  much  as  ylj.  y^-g-  X  13  =  -j9^  of 
a  dollar,  price  of  1  bushel.  Then  -£§  of  y9^  =  ylnb-,  is  tne 
price  of  ^  of  a  bushel,  and  32  times  the  price  of  -j^,  that  is,  yoir 
X  32  =^§^  of  a  dollar,  is  the  price  of  |^§  of  a  bushel. 

NOTE  2.  This  process  is  called  an  analytical  solution,  or  solving 
the  question  by  analysis. 

35.  If  16  men  will  finish  a  piece  of  work  in  28£  days,  how 
'ong  will  it  take  12  men  to  do  the  same  work  ? 

NOTE  3.  Find  in  what  time  1  man  would  do  it,  and  12  men  would 
do  it  in  y1^  of  the  time.  Ans.  37|-  days. 

36.  How  many  yards  of  alpacca,  which  is  l£  yards  wide, 
will  be  required  to  line  20  yards  of  cassimere  £  of  a  yard 
wide  ? 

NOTE  4.  Find  the  square  contents  of  the  cassimere,  which  is  the 
same  as  the  square  contents  of  the  alpacca.  We  have,  then,  the  square 
contents  and  the  width  of  the  alpacca  given  to  find  the  length?  ^]  49, 

Ans.  12  yds.  of  alpacca. 


178  PRACTICE.  IT  135 

37.  If  7  horses  consume  2|  tons  of  hay  in  6  weeks,  how 
much  will  12  horses  consume  in  8  weeks  ?     Ans.  6f  tons. 

38.  If  5  persons  drink  7£  gallons  of  beer  in  1  week,  how 
much  will  8  persons  drink  in  22 J  weeks  ? 

Ans.  28Gf  gallons. 

39.  A  merchant  bought  several  bales  of  velvet,  each  con 
taining  129^4  yards,  at  the  rate  of  $7  for  5  yards,  and  sold 
them  at  $11  for  7  yards,  gaining  $200  on  them;   how  many 
bales  were  there  ?  Ans.  9  bales. 

40.  If  %  Ib.  less  by  |  costs  13|d.,  what  cost  14  Ib.  less  by 
|of21b.?  Ans.  £4  9s.  9<&d. 

41.  If  5  acres  1  rood  produce  26  quarters  2  bushels  of 
wheat,  how  many  acres  will  be  required  to  produce  47  quar 
ters  4  bushels  ?    "  ,  Ans.  9  A.  2  R. 

42.  If  9  students  spend  £10£  in  18  days,  how  much  will 
20  students  spend  in  30  days  ?  Ans.  £39  18s.  4f$d. 

43.  If  f  yd.  cost  $J,  what  will  40J  yds.  cost  ? 

Ans.  $59*062  -|-. 

44.  If  -j3^  of  a  ship  costs  $251,  what  is  ^  of  it  worth  ? 

A?is.  $53*785  +. 

45.  At  £3f  per  cwt.,  what  will  9|  Ibs.  cost? 

Ans.  6s.  3-&d. 

46.  A  merchant,  owning  f  of  a  vessel,  sold  §  of  his  share 
for  $957  ;  what  was  the  vessel  worth  ?       Ans.  $1794<375. 

47.  If  j-  yd.  cost  £f ,  what  will  T9r  of  an  ell  Eng.  cost  ? 

Ans.  17s.  Id.  2fq. 
• 


PRACTICE. 


If  135.     I.   When  the  price  is  an  aliquot  part  of  a  dollar. 
NOTE. — For  the  definition  of  aliquot  part,  see  ^f  55. 

ALIQUOT  PARTS  OF  1  DOLLAR. 

Cents.  Cents. 

50    =  i  of  1  dollar.  12$  =  i  of  1  dollar. 

33£  =  &  of  1  dollar.  10    =  TV  of  1  dollar. 

25    =  £  of  1  dollar.  6£  =  T\,  of  1  dollar. 

20    =  |  of  1  dollar.  5    =  2V  of  1  dollar. 

1 .    What  will  be  the  cost  of  4857  yards  of  calico  at  25 
«ents  (=  £  of  1  dollar)  per  yard  ? 


^  135.  PRACTICE.  •  179 

OPERATION.  SOLUTION.  —  At  $1  a  yard,  the  cost  would 

4)4857  dollars,        be  as  many  dollars  as  there  are  yards,  that  is, 

$4857  ;  and  at  £  cf  a  dollar  a  yard,  it  is  plain, 

$1214'25,   Ans.     that  the  cost  will  be  £  as  many  dollars  as  there 

are  yards,  that  is,  ®4-8^==$1214'2f>. 

After  dividing  the  unit  figure  7,  there  is  a  remainder  of  1,  (dollar,) 
which  we  reduce  to  cents  by  annexing  ciphers,  a»d  continue  the  divi 
sion. 

This  manner  of  computing  the  cost  of  articles,  by  taking 
aliquot  parts,  is  called  PRACTICE,  from  its  daily  use  among 
merchants  and  tradesmen. 

Hence,  when  the  price  is  an  aliquot  part  of  a  dollar,  this 
general 

RULE    OF    PRACTICE. 

Divide  the  price  at  $1  per  pound,  yard,  &c.,  by  the  num 
ber  expressing  the  aliquot  part,  the  quotient  will  be  the  an 
swer  in  dollars. 

NOTE.  —  If  there  be  a  remainder,  it  may  be  reduced  to  cents  and 
mills,  by  annexing  ciphers,  and  the  division  continued. 

EXAMPLES  FOR  PRACTICE. 

2.  What  is  the  value  of  14756  yards  of  cotton  cloth,  at  12J 
cents,  or  £  of  a  dollar  per  yard  ? 

By  practice.  By  multiplication.  NOTE.  —  By 

8)  14756                        14756  comparing     the 

_                         %125  ^wo  °Perati°ns> 

ceiQ/i/i<£n  it  will  be  seen 

$1844-50  —  -  that  the  opera. 

OQ^IO  *i(m   by  Practice 

29ol2  is  much  shorter 

14756  than  the  one  by 

multiplication. 


$1844'500  Am.  as  before. 

3.  What  is  the  cost  of  18745  pounds  of  tea,  at  $'50,  =  J 
dollar,  per  pound  ?  Am.  $9372'50. 

4.  What  is  the  value  of  9366  bushels  of  potatoes,  at  33* 
cents,  or  £  of  a  dollar,  per  bushel  ?       S^JS-  =  $3122,  Ans. 


Questions.  —  T[  135.  What  do  you  understand  by  aliquot  parts? 
What  are  the  aliquot  parts  of  a  dollar  1  When  the  price  of  1  yard,  1 
pound,  &c.,  is  an  aliquot  part  of  a  dollar,  how  may  the  cost  of  any 
quantity  of  that  article  be  found?  What  is  this  manner  of  computing 
called?  Why?  Repeat  the  rule.  What  two  things  are  g'ven,  and 
what  one  is  required,  in  example  3d  ?  —  in  example  4th?  5th?  Oth?  7th  f 
dec. 


180  .   PRACTICE.  IF  136. 

5.  What  is  the  value  of  48240  pounds  of  cheese,  at  $<06£, 
=  T16  of  a  dollar,  per  pound  ?  Ans.  $3015. 

6.  What  cost  4870  oranges,  at  5  cents,  =  fa  of  a  dollar, 
apiece  ?  Ans.  $243'50. 

7.  What  is  the  value  of  151020  bushels  of  apples,  at  20 
cents,  =  I  of  a  dollar,  per  bushel  ?  Ans.  $30204. 

8.  What  will  264  pounds  of  butter  cost,  at  12J  cents  per 
pound  ?  Ans.  $33. 

9.  What  cost  3740  yards  of  cloth,  at  $1'25  per  yard  ? 

4)  $3740  =  cost  at  $1'  per  yard. 
935  =  cost  at  $  '25  per  yard. 

Ans.  $4675  =  cost  at  $1'25  yer  pard. 

10.  What  is  the  cost  of  8460  hats,  at  SM2J  apiece  ? 

at   $1'50  apiece  ?    at   $3'20  apiece  ?    at  $4'06| 

apiece?      Ans.  $9517<50.    $12690.    $27072.    $34368'75. 

1T  136.  U-  To  find  the  value  of  articles  sold  by  the  100, 
or  1000. 

1.  What  is  the  value  of  865  feet  of  timber,  at  $5  per  hun 
dred  ? 

Were  the  price  $5  per  foot,  it 

OPERATION  *s  P^i11  tne  value  would  be  865 

X  $.r:  -—  $4325  ;  but  the  price  is 
$>5  for  100  feet :  consequently, 
$4325  is  100  times  the  true  value 
of  the  timber ;  and  therefore,  if 

$4325  =  value  at  $5  per  foot.     we  divide  this  number  ($4325) 

by  100,  we  shall  obtain  the  true 
value  ;  which  we  do  by  cutting  off  two  right  hand  figures. 

Ans.  $43'25. 

Were  the  price  so  much  per  thousand,  the  same  remarks  would 
apply,  with  the  exception  of  cutting  off  three  figures,  instead  of  two. 
Hence, 

To  find  the  value  of  articles  sold  by  the  100  or  1000, 

RULE. 

I.    Multiply  the  number  and  price  together. 

11.  If  the  price  be  by  the  100,  cut  off  two  figures  at  the 
right ;  if  by  the  1000,  cut  off  three  figures  at  the  right ;  the 
product  will  be  the  answer,  in  the  same  denomination  as  the 
price,  which,  if  cents  or  mills,  may  be  reduced  to  dollars. 


137.  PRACTICE.  181 

2.    What  is  the  cost  of  4250  bricks,  at  $5'75  per  1000  ? 

Ans.  $24'43f, 

OPERATION. 

$5'75 

4250         In  his  example,  we  cut  off  three  figures  from  the 

*"          right  hand  of  the  product,  because  the  bricks  were 

9R7/5        so^  ky  t^e  100°-     The  remaining  figures  at  the  left 

-  \crJ        express  the  cost  of  the  bricks  in  the  lowest  denomi 

nation  of  the  price,  viz.,  cents,  which  we  reduce  to 

dollars  by  pointing  off  two  places  foi  cents 


$24'43|750 

3.  What  will  3460  feet  of  timber  cost,  at  $4  per 
dred? 

4.  What  will  24650  bricks  cost,  at  5  dollars  per  1000? 

5.  What  will  4750   feet  of  boards   cost,   at    $12'  25  pel 
1000?  Ans.  858487+. 

6.  What  will  38600  bricks  cost,  at  $4'75  per  1000  ? 

7.  What  will  46590  feet  of  boards  cost,  at  S10'625  per 
1000? 

8.  What  will  75  feet  of  timber  cost,  at  $4  per  100  ? 

9.  What  is  the  value  of  4000  bricks,  at  3  dollars  per  1000  ? 

10.  Wilderness,  February  8,  1847. 
Mr.  Peter  Carpenter, 

Bought  of  Asa  Falltree, 

5682  feet  Boards,  at  $6       per  M. 

2000     "         "  "     8;34       " 

800     »     Thick  Stuff,  "  12'64     .  « 

1500     »     Lathing,  "     4 

650     «     Plank,'  "   10 

S79     »     Timber,  "     2'50  per  C. 

236     "  "  "     2'75       " 


Received  payment,  $101 '849. 

Asa  Falltree. 

NOTE. — M.  stands  for  the  Latin  mille,  which  signifies  1000,  and 
C.  for  the  Latin  word  centum,  which  signifies  100. 

IT  1 37.    III.    To  find  the  cost  of  articles  ly  the  ton  of  2000  Ibs. 
1.    What  cost  3684  Ibs.  of  hay,  at  $12'40  a  ton  ? 

Questions.  — ^[  136.     To  find  the  cost  of  articles  sold  by  100.  or 

tOOU,  what  is  the  first  step  prcposed  by  the  rule  ?  —  the  second  ?  In  what 

denomination  will  the  produt  ;  be?     How  will  you  find  the  cost  of  725 

bricks,  at  i4i25  a  thousand?     How  many  figures  in  all  do  we  point  off? 

16 


cu 

,' 


182  PRACTICE.  H  13?. 

OPERATION.  SOLUTION.  —  A  ton 

$12;40  -r-  2= $&2Q= price  of  1000  Ibs.     equals  2000  Ibs.     Di- 

$6'20  viding  the  price  of  one 

ton  by  2,  we  have  the 
price     of     1000  Ibs., 

S22'84I080  which  is  $6<20.    Mul 

tiplying   this   cost   by 
the  number  of  pounds, 

3684,  it  will  give  the  cost  at  $6'20  per  pound,  a  result,  which  is  1000 
times  too  large.  If  136.  We  therefore  divide  this  product  by  1000, 
cutting  off  three  right  hand  figures,  and  have  the  cost  at  $6'20  per 
'000  iks.,  or  $12'40  per  ton. 

Ans.  $22'84+. 
Hence, 

RULE. 

Multiply  1  half  the  cost  of  1  ton  by  the  number  of  pounds, 
and  point  off  three  figures  from  the  right  hand.  The  remain 
ing  figures  will  be  the  price,  in  the  denomination  of  the  price 
of  1  ton,  which,  if  cents  or  mills,  may  be  reduced  to  dollars. 

NOTE.  —  At  $12'40  per  ton,  or  $6'20  per  1000  Ibs. 
100  Ibs.  will  cost  $  '62  removing  the  separatrix  1  figure  to  the  left. 
10  "         "         $'062        "  "  2  figures        " 

1  "         "         $'0062      "  "  3       "  " 

2.  What  is  the  cost  of  15742  Ibs.  of  Anthracite  coal,  at 
$7'50  per  ton  ?  Ans.  $59'032-f-. 

3.  What  will  be  the  transportation  on  49826  Ibs.  of  iron, 
from  New  York  to  Chicago,  at  $11  per  ton  ? 

Ans.  $274'043. 

4.  What  will  be  the  storage  on   13991  Ibs.  of  goods,  at 
$2'50  per  ton  ?  Ans.  $17<488+. 

5.  What  will  be  the  cost  of  658  Ibs.  of  hay,  at  S7'38  per 

ton  ?  at  $5'25  ?  at  $8'50  ?  at  $9'00  ?  at 

$9'50?  at  $11  ? at  $12? 

Ans.  $2'428.  $1*727.  $2'796L  $2'961.  $3<125J.  $3'619. 
S3'948. 

6.  At  $7'00  per  ton,  what  will  be  the  cost  of  424  Ibs.  of 

hay?  530  Ibs.?  658  Ibs.?  750  Ibs.?  896 

Ibs.?   —  918  Ibs.?   1024  Ibs.?    1216  Ibs.?   

1350  Ibs.  ?   1600  Ibs.?   1890  Ibs.  ? 

Ans.  $1'484.  $l'85i.  $2>303.  $2-62J.  $3' 136.  $3'213, 
$3*584.  $4'256.  $4'72J.  $5'60.  $6l61i. 


Questions.  —  ^[  137.  When  the  cost  of  1  ton  is  given,  how  do  you 
find  the  cost  of  1000  pounds  ?  How  do  you  find  the  cost  of  any  number 
of  pounds  ? 


IT  138.  PRACTICE.  188 

1T  138.     IV.    When  the  price  is  the  aliquot  part  of  a  £. 

.ALIQUOT    PARTS    OF    A   £. 

of  1£.  6s.  8d.  =  1  of  1£. 

"  3s.  4d.  =  £      " 

"  2s.  6d.  ==-  I      " 

"  Is.  8d.  =  TV     " 

1.  Bought  of  John  Smith,  Liverpool,  7685  yards  of  black 
broadcloth  at  10s.  per  yard ;  what  did  it  cost  ? 

OPERATION.  SOLUTION.  —  Atl<£.per 

2)7685£.  =price  at  1£.  per  yard,     yard  the  price  is  7685^., 

and  one  half  of  this  is  the 

3842£.  105.  =  "      105.        «  price  at  10s  .per  yard.    The 

remainder  of  \£.  may  be 
reduced  to  shillings  and  then  divided. 

Hence,  when  the  price  is  the  aliquot  part  of  !£., 

RULE. 

Divide  the  price  of  the  quantity  at  1£.  per  yard,  bushel, 
&c.,  by  the  number  expressing  the  aliquot  part.  The  answer 
will  be  in  pounds. 

EXAMPLES    FOR    PRACTICE. 

2.  What  cost  1873  reams  of  paper,  at  6s.  8d.  per  ream  ? 

Ans.  624£.  6s.  8d. 

3.  What  cost  10416  bushels  of  salt,  at  3s.  4d.  per  bushe.  ? 

Ans.  1736£. 

4.  Bought  640  Ibs.  colored  thread,  at  7s.  6d.  per  pound; 
what  was  the  whole  cost  ? 

OPERATION.  SOLUTION.  —  7s.  6d.  is  not  an  ali- 

8  \  4)  640£   ==  1£   ver  Ib.  <luot  Part  of  a  pound,  but  it  is  equal  to 

' '  5s. -f  2s.  6d.,  and  taking  k  of  640 JC. 

91       IfiO-P ^            «  we  have  the  price  at  5s.  per  Ib.,  and 

1         «n?'~     9     (U  <i  iof640^.,oriofl60jC.,istheprice 

BUi.  =  Zs.  ba.  at  2Se  6di     Then5  adding  together  the 
prices  at  5s.  and  2s.  6d.  per  Ib.,  we 

240£.  =  75.  6d.  "  have  the  price  at  7s.  6d.  per  Ib. 

5.  What  cost  866  yards  of  black  silk,  at  14s.  per  yd.  ? 

A7is.  606£.  4s. 

6.  What  cost  7  T.  8  cwt.  of  iron,  at  16s.  8d.  per  cwt.  ? 

Ans.  123£.  6s.  8d. 


Questions.  —  If  138.  Give  tne  aliquot  parts  of  1£.  Explain  the 
principle  on  which  the  first  example  is  performed  ?  —  the  fourth  exam 
ple?  Rule. 


184  PRACI  ICE.  T  139,  140. 

^j"  139*     V.    When  the  price  is  an  aliquot  part  of  a  shilling. 

ALIQUOT    PARTS    OF    A    SHILLING. 

6d.  =  J  of  Is.  2d.    =  £  of  Is. 

4d.  =  i     "  1  Jd.  =  i     " 

3d.  ==|     "  Id.    =TV     " 

Reasoning  as  above,  we  have  this 

RULE. 

Divide  the  price  of  the  quantity  at  Is.  per  lb.,  yd.,  &c.,  by 
the  number  expressing  the  aliquot  part;  the  answer  will  be 
in  shillings. 

EXAMPLES  FOR  PRACTICE. 

1.  Sold  348216  Ibs.  of  cotton,  for  4d.  per  lb.  ;  what  did  1 
receive  ?  Ans.  116072s.  =  5803£.  12s. 

2.  Bought  2490  yds.  of  calico,  at  9d.  per  yard  ;  what  did  it 
all  cost  ?     9d.  =6d.  +  3d.  Ans.  93£.  7s.  6d. 

3.  Bought  4000  papers  of  pins,  at  4£d.  per  paper  ;  what 
was  the  cost  ?     4Jd.  =  3d.  -|-  1  Jd. 

4.  What  cost  7430  Ibs.  of  sugar,  at  6d.  per  lb.  ?   -  at 
4d.  ?  -  at  3d.  ?  -  at  2d.  ?  -  at  1  Jd.  ?  -  at  Id.  ? 

IT  1  4LO.  To  find  the  price  of  a  quantity  less  than  unity, 
when  it  n  an  aliquot  part,  or  parts,  of  I. 

1.  At  SI  '50  per  bushel  for  wheat,  what  will  2  pecks  and  4 
quarts  cost? 


ft  !  sn'          SOLUTION.  —  We  divide  the  price  of  1  bushel  by  2, 
which  gives  the  price  of  2  pecks,  or  i  a  bushel.    Then, 
as  4  quarts  is  £  of  1  bushel,  we  divide  $1'50  hv  8,  or 
4  )    '75        as  it  is  £  of  half  a  bushel,  we  divide  $'75  by  4,  for  the 
'1S|      price  of  4  quarts,  and  the  quotient,  added  to  the  price  of 
_     2  pecks,  gives  the  price  of  2  pecks  and  4  quarts. 
<93|  Ans.  *<93|. 

Hence, 

RULE. 

Take  such  part,  or  parts,  of  the  price  of  unity  as  the  quan 
tity  is  of  1  ;  the  part,  or  sum  of  the  parts  taken,  will  be  the 
price  of  the  quantity. 

Questions.  —  ^[  139.  Give  the  aliquot  parts  of  1  shilling.  Rule 
Show  how  the  2d  example  can  be  performed  by  Practice  ;  the  3d. 

1J  140.  What  is  the  subject  of  this  If  ?  Why  divide  $1<50  by  8,  to 
get  the  pme  of  4  qts.  of  wheat?  Why  divide  S'75  by  4  for  the  sain* 
purpose  ?  Give  th«  rule. 


H  141.  PRACTICE.  185 

EXAMPLES  FOR  PRACTICE. 

2.  What  costs  3  qts.  of  oil,  at  $494  per  gal.  ? 

Am.  8'70i. 

3.  What  shall  I  receive  for  building  90  rods  of  road,  at 
11200  per  mile  ?  Am.  $337'50. 

4.  Bought  65  Ibs.  of  pork,  at  $17'25  per  barrel  ;  what  did 
I  pay  ?  •  Am.  $5'60  +. 

5.  What  will  14  quires  of  paper  cost,  at  $3'  00  per  ream  ? 

Ans.  $2'  10. 

6.  At  88-50  per  month  of  30  days  for  the  rent  of  a  house, 
what  will  be  the  rent  for  18  days  ? 

2).  $8'  50  SOLUTION.—  Take  half  of  the  rent  for 

30  days,  which  will  be  the  rent  for  15  days, 

5)  4'25/or  15  days,     and  one  fifth  of  the  rent  for  15  days  will  be 
'85  for    3  days,     the  rent  for  3  days,  and  add  together  the 
_  rent  for  15  and  3  days,  the  sum  will  be  the 

rent  for  18  days. 


7.  What  will  a  man's  salary  amount  to  in  7  months,  at  the 
rate  of  $500  a  year  ? 

2)  $500 

6)  ~250  =  /br  6  months.        SOLUTION.  -One  half  the  year's  salary 
41<£fi_L_    1  W1*l  ')e  tne  salary  f°r  6  months,  and  one 

sixth  of  this  the  salary  for  1  month. 

291'66+,  7  mo. 

8.  What  will  be  a  man's  salary  for  8  months  and  21  days, 
at  $400  per  annum,  that  is,  by  the  year  ? 

Ans.  $290. 

9.  What  will  5  cord  feet  and  12  solid  feet  of  wood  cost,  at 
$2'50  per  cord  ?  Ans.  $1'SO,  nearly. 

10.  What  will  11  oz.  of  sugar  cost,  at  12  cents  per  pound? 

Ans.  $'082  J. 

11.  What  will  3f  yards  of  broadcloth  cost,  at  $4'00  peui 
yard?  Ans.  $14'  50. 

IT  141  .  To  reduce  shillings,  pence,  and  farthings,  to  tkt 
decimal  of  a  pound,  by  inspection. 

There  is  a  simple  and  concise  method  of  reducing  shillings, 
pence,  and  farthings  to  the  decimal  of  a  pound,  by  inspection. 
The  reasoning  in  relation  to  it  is  as  follows  : 

-j^j  of  20s.  is  2s.  ;  therefore  every  2s.  is  ^j-,  or  '!£.  Every 
shilling  is  ^  =  ^^  or  '05£.  Pence  are  readily  reduced  tP 
16* 


186  PRACTICE.  H"  141. 

farthings.  Every  farthing  is  •g-J7T£.  Had  it  so  happened 
that  1000  farthings,  instead  of  960,  had  made  a  pound,  then 
every  farthing  would  have  been  TuW>  or  '00 1£.  But  960 
increased  by  ^  part  of  itself  is  1000  ;  consequently,  24  far 
things  are  exactly  1%^,  or  '025£.,  and  48  farthings  are  exactly 
rt$tr>  or  '050£.  For,  add  ^  of  any  number  of  farthings  to 
the  number,  and  it  will  be  reduced  to  thousandths  of  a  pound. 

If  the  farthings  are  20,  they  will  equal  '020f£ ,  which  we 
will  call  '021,  since  fj  is  more  than  J  of  a  thousandth.  If  the 
farthings  are  14,  they  will  equal  '014^J-,  which  we  will  call 
'015  for  the  same  reason.  But  if  the  farthings  are  only  10, 
they  will  equal  '010££,  which  we  call  '010,  since  ££  is  less 
than  \  a  thousandth.  If  the  farthings  are  31=^1^  = 
32/¥,  we  call  them  '032,  for  the  same  reason.  And  if  the 
farthings  be  42  ==  42ff  =  43£f ,  we  call  them  '044.  The  re 
sult  will  always  be  nearer  than  \  of  1  thousandth  of  a  pound. 
Thus,  17s.  5fd.  is  reduced  to  the  decimal  of  a  pound  as  fol 
lows :  16s.  =  '8£.  and  Is.  = '05£.  Then  5fd.  =  23  far 
things,  which,  increased  by  1,  (the  number  being  more  than 
12,  but  not  exceeding  36,)  is  '024£.,  and  the  whole  is  '874£., 
the  Ans. 

Wherefore,  to  reduce  shillings,  pence,  and  farthings  to  the 
decimal  of  a  pound,  by  inspection,  —  Call  every  two  shillings 
one  tenth  of  a  pound  ;  every  odd  shilling,  five  hundredths  ; 
and  the  number  of  farthings,  in  the  given  pence  and  farthings, 
so  many  thousandth,  adding  one,  if  the  number  be  more  than 
twelve  and  not  exceeding  thirty-six,  and  two,  if  the  number  be 
more  than  thirty-six. 

NOTE.  —  If  the  farthings  be  just  12  =  '012^f,  they  are  equal  to 
'0125  ;  if  36,  they  are  equal  to  '0375.  48  farthings  =  12d.  =  Is.  equal 
'05  of  a  pound,  as  above. 

EXAMPLES   FOR    PRACTICE. 

1.  Find,  by  inspection,  the  decimal  expressions  of  9s.  7d., 
and  12s.  0{d.  Am.  '479£.,  and  '603£. 

2.  Reduce  to  decimals,  by  inspection,  the  following  sums, 
and  find  their  amount,  viz. :"  15s.  3d. ;  8s.  11-J-d. ;   10s.  6^d.  ; 
Is.  8|d. ;  Jd.,  and  2Jd.  Amount,  £1'833. 


Questions.  —  IT  141.  What  is  the  rule  for  reducing  shillings, 
pence,  and  farthings,  to  the  decimal  of  a  pound,  by  inspection  ?  What' 
is  the  reasoning  in  relation  to  this  rule? 


fl  142,  143.  PERCENTAGE.  187 


To  reduce  the  decimal  of  a  pound  to  shillings, 
pence,  and  farthings,  by  inspection. 

Reasoning  as  above,  (f[  141,)  the  first  three  figures  in  any 
decimal  of  a  pound  may  readily  be  reduced  to  shillings, 
pence,  and  farthings,  by  inspection.  Double  the  first  figure, 
or  tentlis,  for  shillings,  and  if  the  second  figure,  or  hundredths, 
be  Jive,  or  more  than  five,  reckon  another  shilling  ;  then,  after 
the  five  is  deducted,  call  the  figures  in  the  second  and  third 
places  so  many  farthings,  abating  one  when  they  are  above 
twelve,  and  two  when  above  thirty-six,  and  the  result  will  be 
the  answer,  within  £  a  farthing.  Thus,  to  find  the  value  of 
'876£.  by  inspection  :  — 

<8      tenths  of  a  pound  =16  shillings. 

'05    hundredths  of  a  pound  =   1  shilling. 

'026  thousandths,  abating  1,  =  25  farthings,      =  Os.  6Jd. 

'876  of  a  pound,  =17s.  6£d. 

Ans. 

1.  Find,  by  inspection,  the  value  of  ,£'523,  and  £'694. 

Ans.  10s.  5£d.,  and  13s. 

2.  Find  the  value  of 


NOTE.  —  When  the  decimal  has  but  two  figures,  after  taking  out 
the  shillings,  the  remainder,  to  be  reduced  to  thousandths,  will  re 
quire  a  cipher  to  be  annexed  to  the  right  hand.  Ans.  9s.  4|d. 

3.  Value  the  following  decimals  by  inspection,  and  find 
their  amount,  viz.  :  ,£'785,  ,£'357,  .£'916,  £'74,  £'5,  £'25, 
£'09,  and  £'008.  Ans.  £3  12s.  lid. 


PERCENTAGE. 

fl"  143.     1.  A  man  owns  a  farm  of  320  acres,  5  per  cent. 
of  which  is  marsh  ;  how  many  acres  are  marsh  ? 

OPERATION.         SOLUTION.  —  Per  cent,  signifies  hundredth  part. 
3<r>0  The  number  placed   before  per  cent,  signifies  how 

^E-  many  hundredths  are  taken,  being  really  the  second 

figure  of  a  decimal  fraction  ;  thus,  5  per  cent,  of  320 
acres  is  '05  of  that  quantity,  and  since  of  implies 


Questions.  —  IT  142.  How  may  the  first  three  figures  of  any 
decimal  of  a  pound  be  reduced  to  shillings,  pence,  and  farthings,  by 
inspection  ?  Explain  the  reasons  for  this  operation. 


188  PERCENTAGE.  If  143 

multiplication,  we  multiply  320  by  '05,  pointing  off  as  in  decimal 
fractions,  and  get  the  A ns.  16  acres. 

The  finding  of  a  certain  per  cent,  or  a  certain  number  of 
hundredths  of  a  quantity,  is  called  percentage  ;  and  it  is  per 
formed  by  the  following 

RULE. 

Multiply  the  quantity  by  the  rate  per  cent.,  written  dec. 
mally  as  hundredths. 

NOTE.  —  Per  cent,  is  from  the  Latin,  which  signifies  by  the  hi 
dred. 

7  per  cent,  is  '07.     25  per  cent,  is  '25.     50  per  cent,  is  '50. 

100  per  cent,  is  I'OO  (|££,  or  the  whole.) 

125  per  cent,  is  1'25  (T§£,  more  than  the  whole.) 

1  per  cent,  is          .          .      , ..,.-._      .          .          .         „          '0' 
|  per  cent,  is  a  half  of  1  per  cent.,  (J  of  1  hundredth,  or 

-nnro  of  the  whole,)       .          .          .          .   .     '00* 
J  per  cent,  is  a  fourth  of  1  per  cent,  that  is,  J  of  1  ^,  =  '002A 
j  per  cent,  is  3  times  J  per  cent.,  that  is,  f  of  y^,  =     '0075 
£  per  cent.,  (£  of  a  hundredth,  that  is,  £  of  TT^-,)  =      '0012;». 
4|  per  cent,  is  '04j  =  '045,  (the  5  expressing  lOths  of 

lOOths,)  .      :    ..         .          .     .  y>      '045. 

EXAMPLES. 

Write  2J  per  cent,  as  a  decimal  fraction. 
2  per  cent,  is  '02,  and  J  per  cent,  is  '005.     Ans.  '025. 

Write  4  per  cent,  as  a  decimal  fraction.    4J  per  cent. 

4|  per  cent.    5  per  cent.    7£  per  cent.   

8  per  cent.  8£  per  cent. 9  per  cent.   9£  per 

cent.     10  per  cent.     10 J  per  cent.     12|  per 

cent.  121  per  cent.  133^  per  cent. 

EXAMPLES    FOR    PRACTICE. 

2.  A  farmer  gives  10  per  cent,  of  460  bushels  of  wheat  for 
threshing;  how  many  bushels  does  he  give  ? 

Ans.  46  bushels. 

3.  A  farmer  rented  ground  on  which  409  bushels  of  oats 
were   raised,  receiving  30  per  cent,  for  the  rent ;  how  many 
bushels  did  he  receive  ?  Ans.  122<7  bushels. 

Questions.  —  ^[  143.  What  does  per  cent,  signify? — percentage? 
—  the  number  before  per  cent.  ?  How  is  '05  of  a  quantity  obtained,  and 
why?  Give  the  rule  for  percentage.  How  is  any  per  cent,  from  1  to 
99  expressed  ?  100  per  cent.  125  per  cent.  ?  £  per  cent.  ?  4£  per  cent  ? 


1[  143.  PERCENTAGE.  189 

4.  A  beef  weighs  895  Ibs.,  of  which  9  per  cent,  is  bone; 
what  does  the  meat  weigh  ?  Ans.  814<45  Ibs. 

5.  A  schooner,  freighted  with  725  barrels  of  flour,  encoun 
tered  a  storm,  when  it  was  found  necessary  to  throw  28  per 
cent,  of  the  cargo  overboard ;  how  many  barrels  were  thrown 
overboard,  and  how  many  were  saved  ? 

Ans.  to  the  last,  522  barrels. 

6.  A  forwarding  merchant  agreed  to  transport  2000  bush 
els  of  corn,  worth  $692'75,  from  Buffalo  to  Albany  for  12| 
per  cent,  on  its  value  ;  what  was  the  cost  of  transportation  ? 

Ans.  $86<59f . 

7.  A  farmer  had  a  flock  of  639  sheep,  which  increased  33  J 
per  cent,  in  1  year ;  how  many  sheep  had  he  at  the  expira 
tion  of  the  year  ?  Ans.  852  sheep. 

8.  A  man  owing  a  debt  of  $1942'71J,  pays  16|  per  cent. 
of  it ;  how  much  of  it  remains  due  ?       Ans.  $1624'595  -}-. 

9.  A  man,  worth  $4861,  lost  28  J  per  cent,  of  it  by  en 
dorsing  with  his  neighbor;  how  much  of  it  did  he  lose  ? 

Ans.  81385*386. 

10.  What  is  |  per  cent,  of  $115?  Ans.  $'862J. 

11.  What  is  |  per  cent,  of  $376  ?  Ans.  $3'29. 

12.  A  gentleman,  worth  $4280,  spent  15J  per  cent,  of  his 
property  in  educating  his  son;  how  much  did  the  son's  edu 
cation  cost  his  father  ?  Ans.  $663'40. 

13.  A  merchant  has  outstanding  accounts  to  the  amount 
of  $1960 ;  22  per  cent,  of  which  is  due  in  3  months,  and  the 
remainder   in   6   months.     What   is   the  amount  due  in  3 
months  ?  in  6  months  ?       Ans.  to  the  last,  $1528'80. 

14.  A  merchant  who  fails  in  business  pays  63  per  cent,  on 
his   debts;    what  does  a  man  receive   whose  demands  are 
$2465?  Ans.  $1552<95. 

15.  What  does  another  man  lose,  whose   demands   are 
$3615  against  the  same  merchant  ?  Ans.  $1337'55. 

16.  A  young  man  is  left  with  $5000,  and   loses  15  per 
cent,  in  paying  too  high  a  price  for  a  farm,  15  per  cent,  of  the 
remainder  in  selling  the  farm  for  less  than  its  value  ;  he  ex 
pends  15  per  cent,  of  what  is  left  in  an  excursion  to  the  west, 
15  per  cent,  of  what  he  has  when  he  gets  back  in  an  unfortu 
nate  investment  in  railroad  stocks,  and  15  per  cent,  of  the  resi 
due  in  trade  ;  what  has  he  then  left  ?      Ans.  $2218'526  (-. 

NOTE.  —  Under  the  general  subject  of  Percentage  will  be  consid- 
ered  Insurance,  Stocks,  Brokerage,  Profit  and  Loss,  Interest,  Dis 
count,  Commission,  Bankruptcy,  Partnership,  Banking,  Taxes  and 
Duties. 


190  PERCENTAGE.  IT  144 


Insurance. 

IT  144.  Insurance  is  security  to  individuals  against  loss 
of  property  from  fire,  storms  at  sea,  &c. 

Companies  incorporated  for  the  purpose,  having  a  certain 
capital  to  secure  their  responsibility,  insure  property  at  so 
much  per  cent,  a  year.  When  any  property  insured  is  de 
stroyed  by  the  agent  insured  against,  the  company  pays  to 
the  owner  the  sum  for  which  it  is  insured.  The  sums  paid  by 
the  several  individuals  insured,  make  up  the  losses,  and  pay 
the  company  for  doing  the  business. 

Premium  is  the  sum  paid  for  insurance. 

Policy  is  the  writing  of  agreement. 

An  Underwriter  is  an  insurer,  whether  it  be  an  incor 
porated  company  or  an  individual. 

Insurance  at  sea,  called  Marine  insurance,  is  usually  for  a 
certain  voyage.  It  is  sometimes  effected  by  an  individual ;  it 
is  then  called  out-door  insurance. 

The  rate  per  cent,  of  insurance  is  in  proportion  to  the  risk. 
Property  is  not  insured  for  its  entire  value,  lest  it  should  be 
fraudulently  destroyed. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  annual  insurance  of  $1000  on  an  academy, 
at  \  per  cent  ?  Ans.  $5. 

&.  Insured  $14500  on  a  factory  at  1  j  per  cent,  per  annum; 
what  was  the  premium  ?  Ans.  $253'75. 

3.  What  is  the  premium  for  insuring  $6000  on  a  store  arid 
goods  at  |  per  cent.  ? 

SOLUTION.  —  At  1  per  cent,  the  sum  is  as  many  cents  as  there  are 
dollars,  or  6000  cents,  which  reduced  is  $  60 '00,  and  £  per  cent,  is  | 
of  this,  or,  Ans.  $  45. 

NOTE.  — In  this  manner  the  percentage  on  any  sum  at  1  per  cent, 
or  less  may  be  calculated  with  ease. 

4.  What  must  be  paid  for  insuring  $800  on  a  farm  house, 
at  %  per  cent.  ?  Ans.  $2. 

Questions.  —  If  144.  What  is  insurance?  Who  insure?  How 
is  insurance  estimated  ?  Who  pays,  if  the  property  be  destroyed  ?  How 
much?  What  remunerates  the  company?  What  is  premium?  — 
policy  ?  —  an  underwriter  ?  —  marine  insurance  ?  TIow  is  marine  in 
surance  often  effected?  What  is  it  then  called?  What  is  life  insurance, 
and  for  what  purpose  is  it  effected  ?  What  is  said  of  health  insurance  ? 
Example.  Why  is  not  property  insured  for  its  entire  value  ? 


IF  145.  PERCENTAGE.  191 

5.  The  Marine  Insurance  Company  insures  $17500  on  the 
cargo  of  the  ship  Minerva,  from  Boston  to  Constantinople,  at 
2  per  cent. ;  what  is  the  premium  ?  Ans.  $350. 

6.  Amos  Lawrence  insures  $34000  on  the  ship  Washing 
ton  and  cargo  from  Canton  to  Boston,  at  8£  per  cent.  ;  v/hat 
does  he  receive  ?  Ans.  $2805. 

7.  The  New  England  Life  Insurance  Company  insures 
$2000  on  a  person's  life  for  one  year  at  a  premium  of  1^  per 
cent.;  what  is  the  sum?  Ans.  $21. 

NOTE  1.  — Life  insurance  is  effected  that  the  heirs  of  the  individual, 
in  case  of  his  death,  may  receive  the  sum  on  which  the  premium  is 
paid.  The  insurance  is  usually  for  one  year,  for  seven  years,  or  for 
life,  and  the  annual  rate  per  cent,  is  determined  by  a  careful  estimate 
made  from  bills  of  mortality  of  the  probable  chances  of  death  with  per 
sons  of  different  ages. 

NOTE  2. — The  premiums  of  health  insurance  companies,  which 
have  lately  been  organized,  are  specified  sums  to  be  paid  annually  ?n 
proportion  to  what  is  received  weekly  in  case  of  sickness.  Thus  in 
the  Massachusetts  Company,  $5'00  a  year  is  paid  by  a  person  at  the 
age  of  30,  to  secure  $4kOO  a  week  in  sickness.  The  premium  it  de 
termined  from  a  careful  estimate  of  the  probabilities  of  health. 

Mutual  Insurance. 

^T  145.  The  rate  at  which  companies  can  afford  to  injure 
is  estimated  from  the  probable  losses  that  will  occur.  But 
when  the  losses  are  small,  large  profits  are  made  by  the  com 
pany.  Or  the  losses  at  some  time  may  be  greater  than  the 
means  at  the  command  of  the  company,  whereby  its  capital 
will  be  annihilated,  while  the  losses  of  the  insured  will  not 
be  fully  made  up. 

Hence,  mutual  insurance  companies  have  been  formed,  to 
average  the  losses  that  may  actually  occur,  which  it  is  the 
aim  of  all  insurance  to  do.  Each  one  gives  a  premium  note 
of  so  much  per  cent,  on  the  property  which  he  wishes  to 
insure,  the  rate  being  determined  by  the  risk  of  the  property. 
The  amount  of  these  notes  are  the  capital  of  the  company, 
and  a  per  cent,  is  paid  down  on  themI  to  furnish  money  for 

Questions.  —  ^[  145.  How  is  the  rate  of  insurance  in  ordinary 
companies  estimated?  What  objections  to  this  method?  What  is  the  aim 
of  all  insurance  ?  Describe  the  premium  note.  How  is  the  rate  deter 
mined?  How  is  money  procured  for  use?  What  is  the  capital  of  the 
company  ?  Why,  and  on  what,  are  assessments  made  ?  Apply  these 
principles  to  Ex.  1,  and  its  solution. 


10»>  PERCENTAGE.  IF  145 

immediate  use.     Any  losses  that  occur  more  than  this  are 
averaged  on  the  premium  notes. 

EXAMPLES    FOR    PRACTICE. 

1.  What  sum  is  paid  by  a  farmer  for  insuring  $1500  on 
his  buildings  for  five  years  in  the  Cheshire  Mutual  Insurance 
Company,  the  premium  note  being  7  per  cent.,  of  which  3 
per  cent,  is  paid  down,  and  assessments  paid  afterwards  of  2, 
1J,  3£,  and  f  per  cent.  ? 

OPERATION. 

$1500 
'07 

105'00  SOLUTION.  —  First  find  the  amount  of  the  premium 

qO1  note,  which  is  7  per  cent,  of  $  1500  =  $  105,  and  3  -\- 

_  2+ 14  +  3|  +  1  =  10i  per  cent,  of  $  105=  11'02£, 


52£         Ans' 
1050 

$11'02J 

2.  What  sum  must  be  paid  for  insuring  $2845  on  a  store 
for  the  same  time,  and  with  the  same  assessments,  the  pro 
mium  note  being  12  per  cent.  ?  A?is.  $35'847. 

3.  What  must  be  paid  as  above,  premium  note   15  per 
cent.  ?  Ans.  $44'808J. 

4.  Insured  $5000  on  a  flouring  mill  for  five  years,  in  the 
Tompkins  Co.  Mutual,  premium  note  22  per  cent,  of  which 
4|  per  cent,  was  paid  down,  and  2^  per  cent,  in  assessments ; 
what  did  it  cost  per  year  ?  Ans.  $15' 40. 

5.  Insured  for  five  years  $3200  on  a  house  of  worship  in 
the  Vt.  Mutual,  premium  note  11  per  cent.,  on  which  3|  per 
cent,  was  paid  down,  and  four  assessments  were  made  re 
spectively  of  1J,  2J,  2,  and  -|  per  cent. ;  what  is  the  whole 
sum  paid?  Ans.  $35'49J 

6.  How  much  more  would  it  cost  to  insure  the  same  prop 
erty  for  the  same  time  in  the  ^Etna  Insurance  Company  at  J 
per  cent,  each  year?  Ans.  $44;50|. 

7.  What  must  be  paid  annually  to  insure  $750  for  five-; 
years  on  a  library,  premium  note  6  per  cent.,  paid  down  4 

.per  cent.,  sum  of  assessments  9-|  per  cent.  ?     Ans.  $1'21J. 

8.  Insured  for  five  years  $900  on  furniture,  premium  note 
5  per  cent.,  sum  of  payments  on  it  6  per  cent. ;  how  much  is 
paid :  Ans.  $2'70. 


«[[  146.  PERCENTAGE.  193 


Stocks. 

IT  14MJ.  In  the  construction  of  a  railroad,  which  costs 
say  $200000,  the  sum  is  divided  into  shares  usually,  of  $100, 
each  individual  paying  the  amount  of  a  certain  number  of 
shares,  which  are  called  his  stock  in  the  road.  The  one  thus 
paying  towards  the  road  is  called  a  stockholder,  and  is  remu 
nerated  by  a  proportional  share  of  the  profits.  It  follows,  of 
course,  that  the  road  may  be  so  profitable  that  each  share  will 
be  worth  more  than  $100 ;  the  stock  is  then  said  to  be  above 
par.  If  the  road  is  unprofitable,  a  share  will  be  worth  less 
than  $100,  and  the  stock  is  said  to  be  below  par.  When  a 
share  is  worth  just  $100,  the  stock  is  said  to  be  at  par.  The 
stockholders  together  constitute  the  railroad  company,  and  the 
sum  of  the  shares  is  the  capital  of  the  company. 

Manufactories,  too  large  for  individual  enterprise,  banks, 
&c.,  are  conducted  in  a  similar  manner. 

When  governments  borrow  money,  the  sum  each  lends  is 
said  to  be  his  stock  in  what  are  called  the  government  funds. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  value  of  35  shares  in  the  Fitchburg  rail 
road,  at  120  per  cent.  ?     f§&  of  $3500  =  how-  much  ? 

Am.  $4200. 

2.  Sold  15  shares  of  the  Eastern  railroad  at  7|  per  cent, 
advance  ;  what  sum  did  I  receive  ?  Ans.  $1612'50. 

3.  What  do  I  pay  for  20  shares   in  the   Old  Colony  rail 
road,  at  1J  per  cent,  below  par?  Ans.  $1975. 

4.  What  are  28  shares  in  the  Vt.  Central  railroad  worth, 
it  11 J  per  cent,  below  par  ?  Ans.  $2485. 

5.  What  are  45  shares  in  the  Exchange  Bank  worth,  at  8 
per  cent,  below  par  ?  Ans.  $4140. 

6.  What  are  30  shares  in  the  Western  railroad  worth,  at 
9J  per  cent,  above  par  ?  Ans.  $3285. 

7.  For  what  must  I  sell  $5000  U.  S.  6  per  cent,  stock, 
that  is,  stock  on  which  6  per  cent,  per  annum  is.  to  be  paid,  at 
1J  per  cent,  discount  ?  Ans.  $4925. 

8.  What  is  $3200  in  the  Amoskeag  Cotton  Manufacturing 
Co.  worth,  at  17  per  cent,  above  par  ?  Ans.  $3744. 

Questions.  —  ^[  14N5.  How  is  a  railroad  built  ?  What  is  a  share  ? 
—  a  stockholder,  and  how  remunerated  ?  When,  and  why,  is  stock  above 
par  ?  —  below  par  ?  —  at  par  ?  What  is  the  company  ?  -  he  capital  ? 
What  else  are  established  on  the  same  plan  ?  What  are  "overnment 
funds  ? 

17 


194  PERCENTAGE.  IN  47,  148. 

9.  What  is  $2000  in  the  Ocean  Steam  Navigation  Co. 
worth,  at  2  per  cent,  advance  ?  Ans.  $2040. 

10.  Bought  9  shares  in  the  Western  Transportation  Co., 
at  4  per  cent,  below  par ;  what  did  I  pay  ?  $864. 


Brokerage. 

IT  147.  Brokerage  is  an  allowance  made  to  a  dealer  in 
money,  stocks,  &c.,  who  is  called  a  broker.  The  allowance 
is  generally  a  certain  per  cent,  of  the  money  paid  out  or  re 
ceived. 

EXAMPLES    FOR    PRACTICE. 

1.  A  western  merchant  procures  83500  in  bills  on  N.  E. 
banks  of  a  Boston  broker,  for  Ohio  money,  the  broker  charg 
ing  2  per  cent,  for  the  accommodation ;  what  brokerage  does 
he  pay  ?  Ans  $70. 

2.  A  drover  exchanges  $2240  of  country  money  for  city 
bills,  paying  %  Per  cent,  on  his  country  money  ;  what  does  he 
receive?  Ans.    $2237'20. 

3.  A  broker  is  directed  to  buy  150  shares   of  the  N.  Y. 
and  Erie  railroad  stock.     He  pays  $92  per  share,  and  re 
ceives  j  per  cent,  on  the  money  advanced ;   what  does  he 
receive  on  the  whole  ?  Ans.  S103'50. 

4.  What  must  I  pay  a  New  York  broker  for  $5000  of  city 
bills,  in  bills  on  eastern  banks,  at  4  per  cent.  ? 

Ans.  $5012<50. 

5.  A  broker  sells  for  an  individual  90  shares  of  the  Fitch- 
burg  railroad  for  $125  a  share,  receiving  1  per  cent,  on  what 
money  he  gets  ;  what  does  he  receive  ?          Ans.  $112>5Q. 

6.  Bought  $6000  in  gold   coin,  paying  the  broker  1  per 
cent,  for  it ;  what  does  he  receive  ?  A?is.  $60. 

7.  Sold  $5200  in  gold  sovereigns,  at  a  discount  of  J  per 
cent.,  for  good  bank  bills,  which  are  more  convenient  for  me 
to  carry;  how  much  in  bills  do  I  receive  ?        Ans.  $5174. 


Profit  and  Loss. 

*R  148.     1.  Bought  cloth  at  40  cents  a  yard;  how  must 
I  sell  it  to  gain  25  per  cent.  ? 

Questions.  —  If  147.     What  is  brokerage  ?  — a  broker?    How  is 
orokerage  calculated? 


f  149.  PERCEjVlAGfc.  195 

SOLUTION.  —  When  the  price  at  which  goods  are  bought  is  given 
to  find  the  price  for  which  they  must  be  sold,  in  order  to  gain  or  lose  a 
certain  per  cent.,  the  calculation  is  by  the  general  rule  for  percentage. 

Ans.  $'50. 

NOTE.  —  The  profit  or  loss  must  be  added  to  or  subtracted  from 
the  price  of  purchase. 

EXAMPLES    FOR    PRACTICE. 

2.  Bought  a  hogshead  of  molasses  for  $60 ;  for  how  much 
must  I  sell  it  to  gain  20  per  cent.  ?  Ans.  $72. 

3.  Bought  broadcloth  at  $2,50  per  yard;  but,  it  being 
damaged,  I  am  willing  to  sell  it  so  as  to  lose  12  per  cent. ; 
how  much  will  it  be  per  yard  ?  Ans.  $2'20. 

4.  Bought  calico  at  20  cents  per  yard ;  how  must  I  sell  it 

to  gain  5  per  cent.  ?  10  per  cent.  ?  15  per  cent.  ? 

to  lose  20  per  cent.  ? 

Ans.  to  the  last,  16  cents,  per  yard. 


Interest. 

^T  149.  Interest  is  an  allowance  made  by  a  debtor  to  a 
creditor  for  the  use  of  money. 

Per  annum  signifies  for  a  year. 

The  rate  per  cent,  per  annum  is  the  number  of  dollars  paid 
for  the  use  of  100  dollars,  or  the  number  of  cents  for  the  use 
of  100  cents  for  1  year. 

NOTE.  —  Percentage  has  hitherto  been  computed  at  a  certain  per 
cent.,  usually  without  regard  to  time;  interest  is  computed  at  a  certain 
per  cent,  for  one  year,  and  in  the  same  proportion  for  a  longer  or 
shorter  time. 

Principal  is  the  money  due,  for  which  interest  is  paid. 
Amount  is  the  sum  of  the  principal  and  interest. 
Legal  interest  is  the  rate  per  cent,  established  by  law. 
Usury  is  any  rate  per  cent,  higher  than  the  legal  rate. 
The  legal  rate  per  cent,  varies  in  different  countries,  and 
in  the  different  States.     It  is 

Questions.  —  ^f  148.  What  do  you  understand  by  profit  ? — by  loss  ? 
What  are  given,  and  what  required,  in  this  ^  ?  How  found  ? 

H"  149.  What  is  interest?  How  is  it  computed?  What  does  per 
annum  signify  ?  —  rate  per  cent,  per  annum  ?  What  distinction  do  you 
make  between  percentage  arid  interest?  What  is  the  principal?  —  the 
amount?  — legal  interest?  — usury?  What  is  the  legal  rate  per  cent, 
in  each  of  the  states  ?  When  no  rate  per  cent,  is  mentioned,  what  rate 
per  cent,  is  understood ?  How  is  interest  for  one  year  computed?  —  for 
more  than  a  year  ? 


196  PERCENTAGE.  1f  150 

6  per  cent,  in  all  the  New  England  States,  in  Pennsylva 
nia  Delaware,  Maryland,  Virginia,  N.  Carolina,  Tennessee, 
Kentucky,  Ohio,  Indiana,  Illinois,  Missouri,  Arkansas,  New 
Jersey,  District  of  Columbia,  and  on  debts  or  judgments  in 
favor  of  the  U.  States. 

7  per  cent,  in  New  York,  S.  Carolina,  Michigan,  Wiscon 
sin,  and  Iowa. 

8  per  cent,  in  Georgia,  Alabama,  Florida,  Texas,  and  Mis 
sissippi. 

5  per  cent,  in  Louisiana. 

When  no  rate  per  cent,  is  named,  the  legal  rate  per  cent, 
of  the  state  where  the  business  is  transacted,  is  always  under 
stood. 

At  6  per  cent.,  a  sum  equal  to  ^  of  the  principal  lent  or 
due  is  paid  for  the  use  of  it  one  year;  at  7  per  cent.,  a  sum 
equal  to  y^  of  it,  and  so  of  any  other  rate  per  cent.  Hence, 

To  find  the  interest  of  any  sum  for  1  year,  is  to  take  such 
a  fractional  part  of  the  principal  as  is  indicated  by  the  rate 
per  cent.,  as  in  percentage,  and  by  the  same  rule ;  that  is, 
we  multiply  the  principal  by  the  rate  per  cent. 

For  more  years  than  1,  multiply  the  interest  for  1  year  by 
the  number  of  years. 

Interest  is  computed  not  only  for  one  or  more  years,  but  in 
"  the  same  proportion  "  for  months  and  days. 

IT  15O.  To  find  the  interest  on  any  sum  for  months  at 
any  rate  per  cent. 

1.   What  is  the  interest  of  $216'80,  at  7  per  cent.,  for  1 

month  ?  for  2  months  ?  3  mo.  ?  4  mo.  ?  

5  mo.  ?  6  mo.  ?  7  mo.  ?  8  mo.  ?  9  mo.  ? 

10  mo.?  11  mo.? 

First,  find  the  interest  for  1  year,  and  then  for  the  months 
take  fractional  parts  of  the  interest  for  1  year. 

NOTE.  — The  pupil  will  readily  perceive 

OPERATION.  methods  of  simplifying  and  shortening  the 

$216'80  Principal,     operation,  according  to  ^[  140.     Thus,  he 
'07  rate  per  ct.     may  take  &  of  the  interest  for  1  year,  that 
is,  the  int.  for  6  months,  and  J  the  int.  for 
6  months,  =  the  interest  for  2  months,  and 
add  these  together  for  the  interest  8  months. 

Questions.  —  H"  150.  How  is  interest  calculated  for  months,? 
What  part  or  parts  do  we  take  for  the  interest  3  months  ?  —  4  months  ? 
—  5  months  ?  —  6  months  ?  —  7  months  ?  —  8  months  ?  —  9  months  ? 
fee. 


f  151.                                   PERCENTAGE.  197 

$15'  1760  Int.  12  mo. 

For  1  mo.  take  ^  of  the  int.  for  12  mo.  =  $1'264  -f-  Am. 

"    2  mo.    "     J                  "                 =    2<529-f-  " 

"    3  mo.    "|                 "                 =    3'794  " 

"    4  mo.    "     J                 "                 =    5<058-|-  " 

"    5  mo.    «'  JL                "                 =    6'323-j-  " 

"    6  mo.    "                        "                 =    7<5S8  " 


"  =    S'8524- 

«  =  10*117  + 

"  =  11'3S2 

=  12'6464- 


"    7  mo.  « 

"    8  mo.  « 

"    9  mo.  " 

44  10  mo.  " 

"11  mo.  "  "  =13'911+ 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  interest  of  $450  for  9  months,  at  the  rate 
in  Louisiana  ?     What  is  the  amount  ? 

Ans.  to  the  last,  $466'87£. 

NOTE.  —  The  amount,  which  is  the  sum  due,  is  found  by  adding 
the  principal  and  interest  together. 

3.  What  is  the  amount  of  $S7<50  on  interest  7  months,  at 
the  rate  in  Georgia?  Ans.  $91'583. 

4.  What  will  be  the  interest  of  $163,  for  4  months,  at  the 
rate  in  S.  Carolina?  Ans.  $3'803. 

5.  What  will  be  the  interest  of  $850,  for  10  months,  at  the 
rate  in  Kentucky  ?  Ans.  $42<60. 

^T  151.     To  find  the  interest  on  any  sum  for  days. 

1.  What  is  the  interest  of  $216'80,  at  7  per  cent.,  for  1 
day  ?  -  for  2  days  ?  -  for  3  days,  and  so  on,  to  29 

days? 

NOTE  1  .  —  In  computing  interest,  1  month  is  reckoned  30  days. 

First,  find  the  interest  for  1  year,  then  for  1  month,  as  in 
Ex.  1,  last  IT.  These  operations  we  need  not  repeat  here, 
but  take  the  interest  as  there  found,  $1'264  for  1  month,  of 
which  we  may  take  parts,  thus  : 

Questions.  —  1[  151.  How  many  days  are  called  a  month  in  com 
puting  interest  ?  How  is  the  interest  for  days  found  ?  —  for  3  days  ?  — 
5  days?  —  8  days?  —  10  days?  —  12  days?  —  18  days?  —  21  days? 
—  25  days  ?  —  28  days  ?  Explain  the  principle  of  the  direction  how  to 
take  3^  of  a  number.  What  is  the  amount,  and  how  found? 


198 


PERCENTAGE. 


152. 


For  1  day  take 


5  days 

6 
10 
15 
18 
20 


i 


OPERATION 

3)$1>264       Int.  1  mo. 

of  Int. 

for  1  mo.    =  $'042  - 

_    < 

1  day. 

=    '210- 

1 

5  days 

«              =    '252  - 

_    < 

6     " 

=    '421- 

_    i 

10     » 

=    '632 

15     " 

times  - 

borH"A  =    '758 

18    " 

times  i 

I                   =    '842 

20     " 

* 

"     3 

"     2 

NOTE  2. — To  get  3*5-  of  a  number,  divide  it  by  3,  setting  the 
f/-st  figure  of  the  quotient  1  place  towards  the  right,  as  in  the  opera 
tion.  For  2  days,  take  2  times  the  interest  for  1  day.  For  9  days, 
£  of  int.  for  30  days  plus  ±  of  £  of  int.  for  30  days.  For  11  days, 
\  plus  £.  For  16  days,  take  4  times  the  interest  for  4  days,  or  % 
the  int.  for  1  month,  -\-  the  int.  for  1  day.  For  20  days,  we  may  take 
2  times  the  int.  for  £  of  1  month.  For  25  days,  £  -f-  J  the  int.  for  1 
month,  varying  after  this  manner  as  may  suit  our  convenience. 

EXAMPLES  FOR  PRACTICE. 

2.  What  is  the  interest  of  $400,  at  the  rate  in  Alabama,  for 
9  days  ?  Am.  $'80. 

3.  What  is  the  interest  of  $75,  for  19  days,  at  the  rate  in 
Florida  ?  Ans.  $'31|. 

4.  What  is  the  interest  of  $500,  for  25  days,  at  the  rate  in 
Texas  ?  Ans.  $2'777. 

5F  1«5S.  When  the  time  is  expressed  in  more  than  one  de 
nomination^  as  in  days  and  months,  or  years,  months,  and  days. 

1.  What  is  the  interest  of  $32'25,  for  1  year  7  months 
19  days,  at  4£  per  cent.  ? 

OPERATION. 

$32'25  Principal. 
'045  rate  per  cent. 


12900 

$1*45125  Int.  for  1  year. 
fat.  of  12  mo.  =  '7256  4-      "       6  months. 
6    "    ='1209--    '<;       I  month. 
30  d<z.  =  '0604--      "     15  days. 


15 

3 


='0120-- 

=  tOQ4° 


day- 


Ans.  S2'3741  -f-  Ira*,  for  1  yr.  7  wo.  19  da. 


H  152.  PERCENTAGE.  199 

Hence,   To  find,    Tie  interest  on  any  sum,  for  any  time,  at 
any  rate  per  cent.,  this 

GENERAL,  RULE. 

I.  For  1  year.    Multiply  the  principal  by  the  rate  per  cent., 
pointing  off  as  in  decimal  fractions,  and  the  product  will  be 
the  interest  for  1  year. 

II.  For  more  years  than  1,  multiply  the  interest  for  1  year 
by  the  number  of  years. 

III.  For  months.     First  find  the  interest  for   1  year,  of 
which  take  such  fractional  part  as  is  denoted  by  the  given 
number  of  months. 

IV.  For  days.     Take  such  part  of  the  interest  for  1  month 
as  is  denoted  by  the  given  number  of  days. 

V.  For  years,  months  and  days,  or  for  any  two  of  these  de 
nominations  of  time.     Find  the  interest  of  each  separately, 
and  add  the  results  together. 

EXAMPLES   FOR    PRACTICE. 

1.    What  is  the  interest  of  $84,  for  1  year  9  months  20 
days,  at  the  legal  rate  in  Alabama?  Ans.  $12<133. 

.      2.    What  is  the  interest  of  $147,  for  2  years  8  months  12 
days,  at  the  rate  in  Michigan  ?  Ans.  $27'783. 

3.  What  is  the  interest  of  $248,  for  2  years  6  months  20 
days,  at  9  per  cent.  ?  Ans.  $57'04. 

4.  What  is  the  interest  of  $161'08,  for  11  months  19  days, 
at  the  rate  in  N.  Y.  ?  Ans.  $10'931  +. 

5.  What  is  the  interest  of  $73'25,  for  1  year  9  months  12 
days,  at  8  per  cent.  ?  Ans.  $10'45  -|-. 

6.  What  is  the  interest  of  $910'50,  for  3  years  9  months 
26  days,  at  7  per  cent.  ?  Ans.  $243'609  -f-. 

7.  What  is  the  amount  of  $185'26,  in  2  years  3  months 
11  days,  at  7J  per  cent.  ?  Ans.  $216<947  -f-. 

8.  What  is  the  interest  of  $656  from  Jan.  9  to  Oct.  9  fol 
lowing,  at  |  per  cent.  ? 

tocrt*    n   •     •     7  SOLUTION.  —  Remove 

$656  Principal.  the  separatrix  two  places 

2 )  6'56  =  Int.  1  yr.  at  1  per  cent.    \Q  ^f  le{^  and  theu  sum 

itself  will  express  the  m- 

2)3'28=    "          "          \        "  terest  for  1  year  at  1  per 

cent.,  Tf  149,  \  of  which 

2)1'64=    «     6?no.      «        "  will  be  the  interest  for  1 

'82  =     "     3  mo.      "        "  year  at  £  per  cent  ,  of 

which     take    fractionaj 
3  Int.  9  mo.  at  j.  per  cent.        parts  for  9  months> 

Questions.  —  ^f  152.     Explain  the  operation,  Ex.  1.    What  is  the 
general  rule  ? 


200  PERCENTAGE.  H  153. 

9.  What  is  the  interest  of  $46'28,  for  2  years  3  months 
23  days,  at  5  per  cent.  ?  Ans.  $5'354  -f . 

10.  What  will  be  the  amount  of  $175'25,  in  5  years  8 
months  and  21  days,  at  6  per  cent.  ?         Ans.  $235'448  +. 

11.  What  will  be  the  amount  of  $96'50,  for  1  year  11 
months  29  days,  at  12J-  per  cent.  ?  Ans.  $120'591  +. 

NOTE.  —  At«12£  percent.,  $  of  the  principal  will  be  the  interest 
for  1  year. 

12.  What  is  the   interest  of  $54*81,  for   1   year  and  6 
months,  at  the  rate  in  Louisiana?  Ans.  $4C11. 

13.  What  is  the  interest  of  $500,  for  9  months  9  days,  at 
the  rate  in  Georgia  ?  Ans.  $31. 

14.  What  is  the  interest  of  $62' 12,  for  1  month  20  days, 
at  4  per  cent.  ?  Ans.  $'345. 

15.  What  is  the  interest  of  $85,  for  10  months  15  days,  at 
12 1  per  cent.  ?  Ans.  $9'296. 

16.  What  is  the  amount  of  $53,  at  10  per  cent.,  for  7 
months  ?  Ans.  $56'091. 

17.  What  is  the  interest  of  $327'825,  at  the  rate  in  Flor 
ida,  for  1  year  ?  Ans.  $26'226. 

18.  What  is  the  interest  of  $325,  for  3  years,  at  the  rate 
in  Pennsylvania?  Ans.  $5S'50. 

19.  What  is  the  interest  of  $187'25,  for  1  year  4  months, 
at  the  rate  in  Delaware  ?  Ans.  $14'98. 

20.  What  is  the  interest  of  $694'84,  for  9  months,  at  10 
per  cent.?  Ans.  $52' 113. 

21.  Wh-it  is  the  interest  of  $32' 15,  for  1  year,  at  4J  per 
cent.  ?  Ans.  $1'446  +. 

22.  What  is  the  amount  of  $600,  in  2  years,  at  the  rate  in 
New  England?  Ans.  $672. 

23.  What  is  the  interest  of  $57'78,  for  1  year  4  months  17 
days,  at  4  per  cent.  ?  •  Ans.  $3' 19. 

24.  What  is  the  amount  of  $298'59,  from  May  19th,  1847, 
till  Aug.  llth,  1848,  at  the  rate  in  Texas  ?     (See  IT  127.) 

25.  What  is  the  amount  of  $196,  from  June  14,  1847,  till 
April  29,  1848,  at  5|  per  cent.  ?  Ans.  $205'861  -(-. 

To  compute  interest  for  months  and  days,  when  the  rate  is  6 
per  cent. 

I.    OF  ONE  DOLLAR. 

IT  153.     A  method  brought  out  in  the  "  Scholar's  Arith 
metic, '   1801,   and  revised   in    "Adams'  New  Arithmetic," 


1F  153  PERCENTAGE.  201 

1827,  vill  be  p:eferred  by  many  to  the  foregoing,  in  those 
states  vhere  the  legal  rate  is  6  per  cent. 

Induction.     The  interest  on  $1  for  1  year,  at  6  per  cent., 

oeing 

'06  cents,  is 

'01  cent  for  2  months, 

'005  mills  (or  £  a  cent)  for  1  month  of  30  days,  and 

'001  mill  for  every  6  days ;  6  being  contained  5  times  in  30. 

Hence,  it  is  very  easy  to  find,  by  inspection,  the  interest  of 
1  dollar,  at  6  per  cent.,  for  any  given  time. 

The  cents  will  be  equal  to  half  the  greatest  even  number  of 
months. 

The  mills  will  be  5  for  the  odd  month,  if  there  be  one,  and 
1  for  every  time  6  is  contained  in  the  given  number  of  days, 
with  such 

Part  of  1  mill,  as  the  days  less  than  6,  are  part  of  6  days. 

1.  What  is  the  interest  of  $1,  at  6  per  cent.,  for  9  months 
18  days  ? 

SOLUTION.  — The  greatest  even  number  of  months  is  8,  the  interest 
for  which  will  be  $'04  ;  the  mills,  reckoning  5  for  the  odd  month,  and  3 
for  the  18  (3  times  6  =  18)  days,  will  be  $'008,  which,  added  to  the 
cents,  give  4  cents  8  mills  for  the  interest  of  $1  for  9  months  and  18 
days.  Ans.  $'048. 

2.  What  will  be  the  interest  of  SI  for  5  months  6  days  ? 
6  months  12  days  ?  7  months  ?  8  months  24 


days?   9  months  12  days?    10  months  ?    11 

months  6  days  ?  12  months  18  days  ?  15  months  6 

days  ?  16  months  ? 

3.  What  is  the  interest  of  $1  for  13  months  16  days  ? 

SOLUTION.  —  The  cents  will  be  6,  and  the  mills  5,  for  the  odd 
month,  and  2  for  2  times  6  =  12  days,  and  there  is  a  remainder  of  4 
days,  the  interest  for  which  will  be  such  part  of  1  mill  as  4  days  is 
part  of  6  days,  that  is,  £=  |  of  a  mill.  Ans.  $'067|. 

4.  What  is  the  interest  of  $1  for  12  months  3  days  ? 

Questions.  —  ^[153.  At  6  per  cent.,  what  is  the  interest  of  $1  for 
1  year?  —  for  2  months?  —  for  1  month  of  30  days?  —  for  every  6 
days?  How,  then,  may  the  interest  of  $1,  at  6  per  cent.,  for  any  given 
time,  be  found  by  inspection?  If  there  is  no  odd  month,  and  the  num 
ber  of  days  be  less  than  6,  what  is  to  be  done  ?  Why  ?  The  interest  on 
$1  for  a  number  of  days  less  than  6,  is  what  ?  How  do  you  find  it  writ- 
ren  in  the  examples  which  have  been  given  ? 


202 


PERCENTAGE. 


IT  154. 


NOTE.  —  If  there  is  no  odd  month,  and  the  number  of  days  be  less 
than  6,  so  that  there  are  no  mills,  a  cipher  must  be  put  in  the  place  of 
mills;  thus,  for  12  months  3  days,  the  cents  will  be  '06,  the  mills  0, 
the  3  days  £  a  mill.  Ans.  $'060£. 

5.    What  will  be  the  interest  of  $1,  for  2  months  1  day? 

• 4  months  2  da^  s  ?    6  months  3  days  ?    8 

months  4  days  ?  10  months  5  days  ?  for  3  days  ? 

for  1  day?  for  2  days  ?  for  4  days  ?  for 

5  days  ?  Ans.  to  the  last,  $'000£. 

II.     OF  ANY-  SUM. 

IT  154.  1.  What  is  the  interest  of  $75,  for  10  months 
12  days  ? 

FIRST    OPERATION. 

$'052  Int.  on  $1. 
75 


260 
364 

$3'900  Int.  of  $75. 

SECOND    OPERATION. 

$75 
'052 

150 
375 


SOLUTION.  —  We  find  the  interest  of  $1, 
by  the  last  ^[,  which  is  $'052,  and  75  times 
this  sum,  as  in  the  first  operation,  will  be 
the  interest  of  $75 ;  or,  since  either  factor 
may  be  made  the  multiplicand,  (^|21,)  we 
multiply  $75  by  '052,  thus  taking-  52  thou 
sandths  of  the  principal,  for  that  is  the  part 
taken,  when  the  interest  of  each  dollar  is 
$'052. 


$3'900  Int.  of  $75. 

2.    What  .is  the  interest  of  $56' 13,  for  8  months  5  days  ? 


OPERATION. 

3|2)$56'13  Principal. 


SOLUTION. — 
We  find  the 
interest  of  $5 
for  the  time  to 
be  $'040f,  and 
we  multiply 
the  principal 
by  '040£.  As 
1  thousandth  of 

$2'29197  Int.  for  8  mo.  5  days.  Ans.     the     multipli 
cand  is  taken 

(1  mill  or  thousandth  being-  the  unit  of  the  multiplier)  for  every  6 
days,  for  the  days  less  than  6,  we  take  such  fractional  part  of  the  mul 


224520  Int.  for  8  mo. 
multip.  =2806         "       3  days. 
«       =1871         "      2     » 


IF  154.  PERCENTAGE.  203 

tiplicand  as  the  odd  day  or  days  is  of  6.  Thus,  for  3  days,  we  take 
£  of  the  multiplicand,  which  will  be  the  interest  for  that  time  in  mills. 
For  '2  days,  we  take  £  of  the  multiplicand.  Adding  together  the  in 
terest  for  8  months,  3  days,  and  2  days,  the  sum  will  be  the  interest 
for  8  months  and  5  days. 

NOTE  1.  —  As  the  interest  of  $1  for  6  days  is  1  mill,  that  of  $10  for 
the  same  time  will  be  10  mills  =  1  cent.  Hence,  if  the  sum  on 
which  interest  is  to  be  cast  be  less  than  $10,  the  interest,  for  any  num 
ber  of  days  less  than  6,  will  be  less  than  1  cent ;  consequently,  in  busi 
ness  transactions,  if  the  sum  be  less  than  $10,  such  days  need  not  be 
regarded. 

Hence,  To  find  the  interest  of  any  given  sum,  in  Federal 
Money,  for  any  length  of  time,  -at  6  per  cent., 

RULE. 

I.  Find  the  interest  on  SI  for  the  given  time  by  inspec 
tion. 

II.  Multiply  the  principal  by  this  sum,  written  as  a  deci 
mal,  and  point  off  the  result  as  in  multiplication  of  decimals. 

EXAMPLES    FOR    PRACTICE. 

3.  What  is  the  interest  of  $194,  for  4  months  12  days? 

Ans.  $4'268. 

4.  Interest  of  $263'48,  for  2  mo.  21  days  ?        -    $3'556. 

5.  Amount  of  $985,  for  5  years  8  months  ?      $1319'90. 

6.  Interest  of  $87' 19,  for  1  year  3  months  ?          $6'539. 

7.  "        of  $116'08,  for  11  mo.  19  days  ?          $6'751. 

8.  "        of  &200,  for  8  mo.  4  days  ?  $8' 133. 

9.  "        of  SO'85,  for  19  mo.  ?    "  $'08. 

10.  "  of  $8-50,  for  1  year  9  mo.  12  days  ?     $'909. 

11.  "  of  $675,  for  1  mo.  21  days?  $5'737. 

12.  "  of  $8673,  for  10  days  ?    "  $14<455. 

13.  "  of  $0-73,  for  10  mo.  ?  $'036. 

14.  "  of  $126'46,  for  9  mo.  ?  $5<69. 

15.  "  of  $318',  for  10  mo.  16  days?  $16'748. 

16.  "  of  $418-,  for  1  year  7  mo.  17  days  ?  $40'894. 

17.  "  of  $268'44,foi:3yrs.5mo.26ds.?  $56' 193. 

18.  "  of  $658,  from  Jan.  9  to  Oct.  9  fol 

lowing?  $29<61. 

Questions.  —  Tf  154.  After  the  interest  of  $1  is  found,  how  is  the 
interest  of  $75  found,  by  the  first  operation,  Ex.  1  ?  —  by  the  second 
operation  ?  Why  ?  In  what  denomination  is  the  |-  cf  the  multiplicand 
taken  in  Ex.  2,  and  why  ?  What  is  said  of  the  interest  of  $10  for  less 
than  6  days,  and  way?  Give  the  rule.  How  may  the  interest  of  an1! 
sum  be  found  for  6  cays  ?  —  for  less  than  6  days  ? 


-204  PERCENTAGE.  H  155. 

19.  Interest  of  $96,  for  3  days  ? 

NOTE  2.  —  The  interest  of  $1  for  6  days  being  1  mill,  the  dollars 
themselves  express  the  interest  in  mills  for  six  days,  of  which  we  may 
take  parts. 

20.  '  Interest  of  $73<50,  for  2  days  ? 

21.  "        of  &180'75,  for  5  days? 

22.  "        of  $15000,  for  1  day? 


IT  155.  When  6  per  cent,  interest  is  required  for  a  large 
number  of  years,  it  will  be  more  convenient  to  find  the  inter 
est  for  one  year,  and  multiply  it  by  the  number  of  years  ;  after 
which,  find  the  interest  for  the  months  and  days,  if  any,  as 
usual. 

1.  What  is  the  interest  of  $520'04,  for  30  years  and  6 
months  ? 

OPERATION. 
$520-04  Principal. 
'06 


2)831'2024  Int.  1  year. 
30 


'0720   "  30  years. 
$   15'6012  "     6  mo. 


$951'6732  "  30  years.  6  mo. 

Am.  $9ol'673. 

2.  What  is  the  interest  of  $1000,  for  120  years  ? 

"  Ans.  $7200. 

3.  What  is  the  interest  on  $400  for  10  years  3  months  and 
6  days  ?  Ans.  $246'40. 

4.  What  is  the  interest  of  $220,  for  5  years  ?  for  12 

years  ?  50  years  ?  Ans.  to  the  last,  $660. 

5.  What  is  the  amount  of  $86,  at  interest  7  years  ? 

Ans.  $122' 12. 

6.  What  is  the  amount  of  $750,  on  interest  9  years  4  mo. 
14  days?  Ans.  $1171'75. 

Questions.  —  If  155.     How  do  we  get  6  per  cent,  interest  for  a 
large  number  or  years  ?    H  3w,  when  there  are  also  months  and  days  ? 


If  156,  157.  PERCENTAGE.  206 

To  find  the  interest  on  pounds,  shillings,  and  pence. 

IT  156.  1.  What  is  the  interest  of  £36  9s.  6Jd.,  for  1 
year,  at  6  per  cent.  ? 

Reduce  the  shillings,  pence,  &c.,  to  the  decimal  of  a  pound,  by  in 
spection,  (^[  141,)   then  proceed  in  all  respects  as  in  federal  money 
Having-  found  the  interest,  reverse  the  operation,  and  reduce  the  first 
three  decimals  to  shillino-g,  &c.,  by  inspection,  (^[  142.) 

Ans.  £2  3s.  9d. 

2.  Interest  of  £36  10s.,  for  18  mo.  20  days,  at  6  per  cent.  ? 

Ans.  £3  8s.  l£d. 

3.  Interest  of  £95,  for  9  mo.  ?  Ans.  £4  5s.  6d. 

4.  What  is  the  amount  of  £18  12s.,  at  6  per  cent,  interest, 
for  10  months  3  days  ?  Ans.  £19  10s.  9Jd. 

5.  What  is  the  amount  of  £100,  for  8  years,  at  6  per  cent.  ? 

Ans.  £148. 

6.  What  is  the  amount  of  £400  10s.,  for  18  months,  at  6 
per  cent.  ?  Ans.  £436  10s.  lOd.  3qr. 

7.  What  is  the  amount  of  £640  8s.,  at  interest  for  1  year, 
at  6  per  cent.  ?    -  for  2  years  6  months  ?    -  for  1  0 
years  ?  Ans.  to  the  last,  £1024  12s.  9Jd. 

8.  What  is  the  amount  of  £391  17s.,  for  3  years  3  mo.,  at 
4£  per  cent.  ? 

9.  What  is  the  amount  of  £235  3s.  9d.,  from   March  5, 
1846,  till  Nov.  23,  1846,  at  5J  per  cent,  ?    Ans.  £244  8|d. 

• 

To  calculate  interest  on  notes,  fyc.,  when  partial  payments 
have  been  made. 

IT  157.  Payment  of  part  of  a  note  or  other  obligation,  is 
called  a  partial  payment. 

It  has  been  settled  in  the  Supreme  Court  of  the  U.  States, 
and  their  practice  adopted  by  nearly  all  the  states  in  the 
Union,  that  payments  shall  b<-'  applied  to  keep  down  the  in 
terest,  and  that  neither  interest  nor  payment  shall  ever  draw 
interest.  Hence,  if  the  payment  at  any  time  exceed  the  inter 
est  computed  to  the  same  time,  that  excess  is  taken  from  the 
principal;  but  if  the  payment  be  less  than  the  interest,  the 
principal  remains  unaltered.  Hence,  the 

Questions.  —  ^[  156.  How  do  we  proceed,  when  the  principal  is 
pounds,  shillings,  and  pence  ? 

18 


206  PERCENTAGE.  H  157. 

RULE. 

Compute  the  interest  on  the  principal  to  the  time  when  the 
payment,  or  payments,  (if  the  first  be  less  than  the  interest,) 
shall  equal  or  exceed  the  interest  due  ;  subtract  the  interest 
Troin  the  payment,  or  sum  of  the  payments,  made  within  the 
time  for  which  interest  was  computed,  and  deduct  the  excess 
fiom  the  principal. 

The  remainder  will  form  a  new  principal,  with  which  pro 
ceed  as  with  the  first. 

1.    $116<666. 

Boston,  May  1st,  1842. 

For  value  received,  I  promise  to  pay  James  Conant,  or 
order,  one  hundred  and  sixteen  dollars  sixty-six  cents  and  six 
mills,  on  demand,  with  interest.  Samuel  Rood. 

On  this  note  were  the  following  endorsements  : 
Dec.    25,  1842,  received  $  16'666~| 

July    10,1843,        "         $    1'666  |        NOTE.  —  In  finding  the  times 
Sept.     1,  1844,        "         $    5' 000  }>  for  computing  the  interest,  con- 
June    14,1845,        "         $33'333  |    suit  1  127. 
April  15,  1846,        "         $62'000j 

What  was  due  August  3,  1847  ?  Am.  $23'775. 

NOTE  1.  — The  transaction  being  in  Massachusetts,  the  rate  of  in 
terest  will  be  6  per  cent. 

The  first  principal  on  interest,  from  May  1,  1842,     $116*666 
Payment,  Dec.  25,  1842,  (exceeding  in- 

ferest'due,) $16'666 

Interest  to  time  of  1st  payment,      .         .        4'549 


Remainder  for  a  new  principal,       .         .  .           $104'549 

Payment,  July  10,  1843,  less  than  inter 
est  then  due,  ...  .  Sl'666 

Payment,  Sept.  1,  1844,  less  than  inter 
est  then  due, $5'000 

Amount  carried  forward,  $6*666 


Questions.  —  *JI  157.  What  is  a  partial  payment  of  a  note  or  obli 
gation  ?  What  court  has  established  a  rule  for  computing  interest  on 
notes,  &c..  on  which  partial  payments  have  been  made  ?  How  exten 
sively  is  this  rule  adopted?  Repeat  the  rule.  What  is  the  great  funda 
mental  principle  on  which  this  rule  is  based?  What  is  customary  when 
notes  with  endorseme  its  are  paid  within  one  year  of  the  time  they  are 
given  ? 


If  157.  PERCENTAGE.  207 

Amount  brought  forward,     $6'666 
Payment,  June  14,  1845,        .  .   $33'333 

Amount,  (exceeding  interest  due,)   $39'999 
Interest  from  Dec.  25,  1842,  to  June  14, 

1845,  (29  mo  19  iays,)      .         .         -      15'490 

$80' 040 

Payment,  April  15, 1846,  (exceeding  in 
terest  due,)         $62<000 

Interest  from  June  14,  1845,  to  April  15, 

1846,  (10  mo.  1  day,)         .         .         .        4'015 


Remainder  for  a  new  principal, 
Interest  due  August  3,  1847,  from  April  15,  1846, 
(15  months  18  days,)  .... 


Balance  due  Aug.  3,  1847. 
2.   $867<33. 


Buffalo,  Dec.  8th,  1842. 


On  demand,  for  value  received,  I  promise  to  pay  James 
Hadley,  or  bearer,  eight  hundred  and  sixty-seven  dollars  and 
thirty-three  cents,  with  interest  after  three  months. 

Wm.  R.  Dodge. 

On  this  note  were  the  following  endorsements,  viz. : 
April  16,  1843,  received  $  136'44. 
April  16,  1845,  received  $319'. 
Jan.      1,  1846,  received  $518'68. 

What  remained  due  July  11,  1847?        Am.  $31'765  +. 

3.    $1000. 

Boston,  Jan.  1,  1840. 

For  value  received,  I  promise  to  pay  George  A.  Curtis,  or 
order,  on  demand,  one  thousand  dollars,  with  interest. 

Caleb  Nelson. 

On  this  note  were  the  following  endorsements: — April  1,  1840, 
$24  ;  .Aug.  1,  1840,  $4;  Dec.  1,  1840,  $6  ;  Feb.  1,  1841,  $60  ;  July 
1,  1841  $40;  June  1  1844,  $300 ;  Sept.  1,  1844,  $12;  Jan.  1, 
1845,  $15;  and  Oct.  1  1845,  $50. 

What  remained  due,  June  1,  1846  ?     Am.  $843<OS3  -f. 
NOTE.  2  — Whgn  notes  are  paid  within  one  year  from"  the  time  they 


SOS  PERCENTAGE.  f  158. 

wen.  given,  and  have  endorsements,  it  is  common  to  subtract  from  the 
amount  of  the  principal  for  the  whole  time  the  amount  of  each  en 
dorsement  from  its  date  till  the  day  of  settlement. 

4.    $300. 

Mobile,  (Alabama,)  June  10,  1846. 

For  value  received,  we  jointly  and  severally  promise  to 
Eeuben  Washburn  to  pay  him,  or  order,  on  demand,  three 
hundred  dollars,  with  interest.  Louis  P.  Legg. 

Sanford  Comstock. 

On  this  note  were  endorsements  :  Jan.  2t),  1847,  $116  ;  March  2, 
1847,  $49<50;  April  26,  1847,  $85. 

What  remained  due  June  2,  1847  ? 

NOTE.  — The  rate  in  Alabama  is  8  per  cent.    Ans.  $67'894-f-. 
CONNECTICUT  METHOD. 

51"  158.  The  Supreme  Court  of  Connecticut  have  estab 
lished  a  method  somewhat  different  from  the  TJ.  S.  Court  rule, 
which  may  be  practised  by  those  belonging  to  the  state.  The 
suostance  of  the  method  is  presented  in  the  following 

RULE. 

I.  Payments  a  year,  or  more  than  a  year  after  the  date  of 
the  note,  or  from  each  other,  and  those  less  than  the  interest 
due,  are  treated  according  to  the  U.  S.  Court  rule. 

II.  Find  the  amount  of  any  other  payment  from  date  till 
one  year  from  the  time  the  note  was  given,  or  of  a  former 
cast,  and  subtract  it  from  the  amount  of  the  principal  for  one 
year.     The  remainder  is  a  new  principal. 

NOTE.  —  Tf  the  note  be  settled  in  less  than  a  year  from  the  time  of 
a  cast,  find  the  amount  of  each  subsequent  payment  that  has  been 
made  till  the  time  of  settlement,  and  subtract  it  from  the  amount  of 
the  principal  found  till  the  same  time. 

SHOO. 

Woodstock,  Ct.,  Jan.  1,  1841. 

For  value  received,  I  promise  to  pay  Henry  Bowen,  or 
order,  eleven  hu  idred  dollars,  on  demand,  with  interest. 

James  Marshall. 

On  this  note  are  the  following  endorsements:  —  Sept.  1,  1841, 
$30  :  April  1,  1842,  $200  ;  Dec.  1,  1842,  $  180  ;  March  1,  1844, 
S  195  ;  Sept  16,  1844,  $250  ;  May  16,  1846,  $  100 ;  July  16,  1846,. 
$  170. 

What  remains  due  Jan.  16,  1847?       Ans.  $234434 -(-. 


H  159,  160  PERCENTAGE.  209 

IT  159.  For  calculating  interest  on  a  note  in  Vermont, 
payable  at  a  specified  time,  with  interest  annually,  on  which 
payments  have  been  made  before  it  is  due . 

RULE. 

From  the  amount  of  the  debt  found  till  the  note  is  due,  sub 
tract  the  amount  of  the  payments  with  the  interest  of  each 
from  its  date  till  the  same  time. 

$800 

Townshend,  Vt.,  Sept.  1,  1840. 

For  value  received,  I  promise  W.  R.  Ranney  to  pay  him, 
or  order,  eight  hundred  dollars  in  five  years,  with  interest  * 
annually.  Bushrod  Washington. 

Endorsements:  July  1,  1842,  $200;  Jan.  1,  1844,  $200;  Sept 
1,  1844,  $300. 

What  remained  due   Sept.  1,  1845?  Ans.  $264. 

COMPOUND  INTEREST. 

IT  1OO.  A  promises  to  pay  B  $256  in  3  years,  with  in 
terest  annually ;  but  at  the  end  of  1  year,  not  finding  it  con 
venient  to  pay  the  interest,  he  consents  to  pay  interest  on  the 
interest  from  that  time,  the  same  as  on  the  principal. 

Simple  interest  is  that  which  is  allowed  for  the  principal 
only. 

Compound  interest  is  that  which  is  allowed  for  both  prin-  . 
cipal  and  interest,  when  the  latter  is  not  paid  at  the  time  it 
becomes  due. 

To  calculate  Compound  Interest, 

RULE. 

•  Add  together  the  interest  and  principal  at  the  end  of  each 
year,  and  make  the  amount  the  principal  for  the  next  suc 
ceeding  year.  From  the  last  amount  subtract  the  principal. 

1.  What  is  the  compound  interest  of  $256,  for  3  years,  at 
6  per  cent.  ? 

Questions.  —  T[  16O.   What  is  simple  interest  ?  —  compound  inter 
est  ?    What  is  the  method  >f  computing  compound  interest  ? 
18* 


210  PERCENTAGE.  H  161 

$256  given  sum,  or  first  principal. 
'06 


27T36  amount,  or  principal  for  2d  year. 
'06 

16'2816  compound  interest,  2d  year,  )  added  to- 
71'36       principal,  do.       )    gether. 


287;6416  amount,  or  principal  for  3d  year. 
'06 


17'25346  compound  interest,  3d  year,  )  added  to- 

2S7'641  principal,  do.       }    gether. 

304'899  amount. 

256  first  principal  subtracted. 


Ans.  $4S'S99,  compound  interest  for  3  years. 

2.    At  6  per  cent.,  what  will  be  the  compound  interest,  and 

what  the  amount,  of  $1  for  2  years?   what  the  amount 

for  3  years  ?  for  4  years  ?  for  5  years  ?  for 

6  years  ?  for  7  years  ?  for  8  years  ? 

Ans.  to  the  last,  $1'593  +. 

5F  1O1.  It  is  plain  that  the  amount  of  $2,  for  any  given 
time,  will  be  2  times  as  much  as  the  amount  of  $1 ;  the 
amount  of  $3  will  be  3  times  as  much,  &c. 

Hence,  we  may  form  the  amounts  of  $1,  for  several  years, 
into  a  table  of  ?nultiplicands  for  finding  the  amount  of  any 
sum  for  the  same  time. 

Questions.  —If  161.  How  do  we  compute  by  the  table?  When 
iher<-  ire  months  and  days,  how  do  you  proceed?  In  what  time  will 


IT  161. 


PERCENTAGE. 


TABL.E, 


211 


Showing  the  amount  of  $1,  or  £1,  for  any  number  of  years 
from  1  to  40,  at  5  and  6  per  cent.,  and  from  1  to  20  at  7 
and  8  per  cent. 


5  per  cent. 

6  per  cent. 

Years. 

5  per  cent. 

6  per  cent. 

,•05 

1'06 

21 

2*785963 

3*399564 

1*1025 

14236 

22 

2*925261 

3*603537 

1*15762- 

1*19101- 

23 

3*071524 

3*819750 

1*21550- 

1*26247- 

24 

3*225100 

4*048935 

1'2762S- 

1*33822- 

25 

3*386355 

4*291871 

1*34009- 

1*41851- 

26 

3*555673 

4*549383 

1*40710- 

1*50363- 

27 

3*733456 

4*822346 

1*47745- 

1*59384- 

28 

3*920129 

5*111687 

1*55132- 

1*68947- 

29 

4*116136 

5*418388 

1*62889- 

1*79084- 

30 

4*321942 

5*743491 

1*71033- 

1*89829- 

31 

4*538039 

6*088101 

1*79585- 

2*01219- 

32 

4*764941 

6*453387 

1*88564- 

2*13292- 

33 

5*003189 

6*840590 

1*97993- 

-  !  2*26090- 

34 

5*253348 

7*251025 

2*07892- 

2*39655- 

35 

5*516015 

7*686087 

2*18287- 

2*54035- 

36 

5*791810 

8*147252 

2*29201- 

2*69277- 

37 

6*081407 

8*636087 

2*40661- 

2*85433- 

38 

6*385477 

9*154252 

2*52695 

3*02559- 

39 

6*704751 

9*703507 

2*65329  + 

3*20713- 

40 

7*039989 

10*285718 

7  per  cent. 

8  per  cent. 

Years. 

7  per  cent. 

8  per  cent. 

1*07 

1*08 

11 

2*10485-] 

2*331638- 

1*1449 

1*1664 

12 

2*25219- 

2*518170- 

1*225043 

1*259712 

13 

2*40984- 

2*719623- 

1'3107964- 

1*360488  -) 

14 

2*57853- 

2*917193- 

1*402551  -1- 

1'469328- 

15 

2*75903- 

3*150569- 

1*50073  - 

1*586874- 

16 

2*95216- 

3*402614- 

1*60578- 

1*713824- 

17 

3*15881- 

3*674823- 

1*71818- 

1*850930- 

18 

3*37993  -< 

3*968809- 

1*83845- 

1*999004- 

19 

3*61652- 

4*286314- 

1'9671£- 

2*158924- 

20 

3*86968- 

4*629219- 

NOTE  1.  — When  there  are  months  and  days,  first  find  the  amount 
for  the  years,  and  on  that  amount  cast  the  interest  for  the  months  and 
days ;  this,  added  to  the  amount,  will  give  the  answer. 

1.  What  is  the  amount  of  $600*50,  for  20  years,  at  5  per 
cent,  compound  interest  ?  at  6  per  cent.  ? 


212  PERCENTAGE.  1T  162. 

SOLUTION.  — $  1  at  5  per  cent.,  by  the  table,  is  $2<65329  ;  there 
fore,  2'65329  X  600<50  =  $  1593'30  +  Ans.  at  5  per  cent.  ;  and 
3'20713  X  600'50  =  $  1925'881  -j-  Ans.  at  6  per  cent. 

2.  What  is  the  amount  of  $40'20,  at  6  per  cent,  compound 

interest,  for  4  years  ?  for  10  years  ?  for  18  years  ? 

for  12  years  ?  for  3  years  and  4  months  ?  for 

20  years  6  months  and  IS  days  ? 

Ans.  to  the  last,  $133' 181  +. 

NOTE  2.  — Any  sum  at  6  per  cent,  compound  interest,  will  double 
itself  in  11  years  10  months  and  22  days. 

3.  What  is  the  amount  of  $750,  at  7  per  cent.,  compound 
interest,  for  16  years?  Ans.  $2214' 12. 

4.  What  is  the  amount  of  $150,  at  8  per  cent.,  compound 
interest,  for  20  years  ? 

IT  162.     ANNUAL  INTEREST. 

1.     $500<00. 

Keene,  N.  H.,  Feb.  2d,  1843. 

For  value  received,  I  promise  to  pay  George  Hooper,  or 
order,  five  hundred  dollars,  with  interest  annually  till  paid. 

Henry  Truman. 

What  was  due  Aug.  2,  1847,  no  payment  having  been 
made  ? 

SOLUTION.  —  A  note  like  the  above,  with  the  promise  to  pay  inter 
est  annually,  is  not  considered,  in  courts  of  law,  a  contract  for  any 
thing-  more  than  simple  interest  on  the  principal.  Interest  does  not 
become  principal  by  operation  of  law.  If  the  annual  interest  be  not 
paid,  the  creditor  can  bring  his  action  for  it  at  the  end  of  each  year. 
If  he  neglects  to  do  this  in  Massachusetts,  and  in  some  other  states, 
he  is  considered  as  waiving  his  claim,  and  must  be  contented  to  re 
ceive  simple  interest. 

In  New  Hampshire,  interest  is  allowed  on  the  annual  interest,  in 
the  nature  of  damages  for  its  detention  and  use,  from  the  time  it  be 
comes  payable  till  paid.  Hence,  when  a  note  is  written  "  with  inter 
est  annually,"  the  following  is  the  N.  H. 

COURT    RULE. 

Compute  separately  jthe  interest  on  the  principal,  from  the 
time  the  note  is  given  till  the  time  of  payment,  and  the  inter 
est  on  each  year's  interest  from  the  time  it  should  be  paid,  till 
the  time  of  payment.  The  sum  of  the  interests  thus  obtained 


IT  162.  PERCENTAGE.  213 

will  be  the  interest  sought,  to  which  add  the  principal  for  the 
amount  due. 

Applying  this  rule  to  a  cast  on  the  ahove  note,  the  first 
year's  interest  of  $30,  is  on  interest  3  years  and  6  months ; 
the  2d  year's  interest  is  on  interest  2  years  and  6  months, 
&c. 

The  operation  may  he  written  down  as  follows : 

Interest  on  the  principal,  $500'00,  4  years  6  months,  $135'00 
"  1st  year's  int.  ($30)  3  years  6  months,  6<30 
"  2d  "  "  2  years  6  months,  4'50 

"          3d         "  "  1  year    6  months,        2<70 

«          4th        "  "  6  months,          '90 


Amount  of  interest,  $149<40 
Then  $500  +  $149<40  =  ^i?zs.  $649<40,  amount  due. 

NOTE  1.  —  Among  business  men  the  mutual  understanding  and 
practice,  oftentimes,  is  compound  interest,  when  the  note  is  written 
with  interest  annually ;  but  compound  interest  cannot  be  legally  en 
forced,  unless  it  be  so  expressed  in  the  note.  The  interest,  by  the 
method  presented  in  the  rule,  is  due  at  the  end  of  each  year,  but  as  it 
is  riot  paid  then,  it  is  on  simple  interest,  just  as  any  other  debt  would 
be,  if  not  paid  when  due. 

NOTE  2.  —  The  same  method  is  practised  in  Vermont  when  no  pay 
ments  have  been  made,  and  there  is  no  time  specified  when  the  note 
is  due.  (See  1  159.) 

2.     $1000'00. 

Brattleboro',  Vt.,  June  10,  1842. 

For  value  received,  we,  jointly  and  severally,  promise  to 
pay  Joseph  Steen,  or  order,  one  thousand  dollars,  on  demand, 
with  interest  annually.  Samuel  W.  Ford, 

Stephen  Wise. 
What  was  due  Sept.  10,  1847?  Ans.  $1355*50. 

NOTES.  —  But  when  payments  have  been  made,  we  find  the 
amount  of  the  principal  for  one  year,  and  having  found  the  amount  of 

Questions.  —  H"  162.  When  a  note  is  written  with  the  promise  to 
pay  interest  annually,  how  is  it  considered  in  courts  of  law  ?  If  the 
annual  interest  be  not  paid,  what  may  the  creditor  do?  If  he  neglects 
to  do  this,  how  is  it  considered  by  the  courts  in  Massachusetts,  and  in 
some  other  states?  What  is  the  New  Hampshire  court  practice?  On 
what  principle  is  simple  interest  allowed  on  the  annual  interest  ?  Re 
peat  the  court  rule.  What  mutual  understanding  among  business  men? 
How  can  the  interest  on  interest  be  regarded  as  interest  on  any  debt  ? 
What  other  state  practises  the  same  method  ?  When  payments  have 
been  made,  what  is  done  ? 


214  PERCENTAGE.  IT  163 

each  payment  made  during  that  year  from  its  date  till  the  end  of  the 
same  year,  we  subtract  it  from  that  amount,  and  the  remainder  will 
he  a  new  principal,  with  which  we  proceed  as  before. 

3.      $400. 

Windsor,  Aug.  2,  1844. 

For  value  received,  I  promise  to  Bishop  and  Tracy,  to  pay 
to  them,  or  order,  four  hundred  dollars,  on  demand,  with  an 
nual  interest.  Asa  H.  Truman. 

On  this  note  are  endorsements  as  follows:  April  2,  1845,  $50; 
JuneS,  1845,  $30;  Jan.  2,  1846,  $100;  May  17,  1847,  $80. 

What  remained  due  Aug.  2,  1847  ? 

OPERATION. 

Principal,         .         .     -  •  >         .         .         .  $400<00 

Interest  1  year, 24<00 

1st  payment,  $50 -4- int.  4 mo.  SI  =  $51<00, 
2d        «         $30-fint.2mo.  $'30=30'30.  Ain't,  424<00 

81*30 

New  prin.  342'70  &c. 
Am.  $194*34. 

The  time,  rate  per  cent.,  and  amount  being  given,  to  find  the 
principal. 

IT  163.  1.  What  sum  of  money,  put  at  interest  1  year 
and  4  months,  at  6  per  cent.,  will  amount  to  $61'02? 

SOLUTION.  —  $  1'08  is  the  amount  of  $  1  for  1  year  and  4  months, 
and  $61'02  is  the  amount  of  as  many  dollars  as  the  nurflber  of 
times  $1*08  is  contained  in  $61'02.  $61;024-$  1'08  =  $56'50, 
the  principal  required.  Hence, 

RULE. 

Divide  the  given  amount  by  the  amount  of  1  dollar,  at  the 
given  rate  and  time. 

2.  What  principal,  at  8  per  cent.,  in  1  year  6  months,  will 
amount  to  $8542?  Ans.  $76. 

3.  What  principal,  at  6  per  cent.,  in  11  months  9  days, 
will  amount  to  $99'311  ? 

Questions. — U  163.  What  is  the  subject  of  this  paragraph? 
Repeat  the  first  example.  Why  do  you  divide  &61<02  by  &1<08?  What 
is  the  rule  for  finding  the  principal,  when  the  time,  late  per  cent.,  and 
amount  are  given  ? 


U  164.  PERCENTAGE.  215 

NOTE. — The  interest  of  $1,  for  the  given-  time,  is  '056£  ;  but, 
when  there  are  odd  days,  instead  of  writing-  the  parts  of  a  mill  as  a 
common  fraction,  it  will  be  more  convenient  to  write  them  as  a  deci 
mal,  thus,  '0565  ;  that  is,  extend  the  decimal  to  four  places. 

Ans.  $94. 

4.  A  produce  buyer  purchased  1000  bushels  of  wheat,  on 
credit,  and  agreed  to  pay  15  per  cent,  on  the  purchase  money; 
at  the  expiration  of  4  months  he  paid  the  debt  and  interest, 
which  together  amounted  to  $1500;  what  was  the  value  of 
the  wheat  ?  Ans.  $1428<571  +. 


Discount. 

IT  164.  1.  I  purchase  a  horse,  agreeing  to  pay  for  him 
&106,  one  year  from  the  time  he  comes  into  my  possession, 
without  interest.  When  I  take  the  horse,  I  propose  to  pay 
down,  for  a  just  allowance  ;  what  must  I  pay,  money  being 
worth  6  per  cent.  ? 

SOLUTION.  —  I  should  not  pay  the  whole  sum,  $106;  for  the  one  who 
received  it  might  loan  it  to  a  third  person,  and  receive  $112'36  for  it  at 
the  end  of  the  year,  when  he  should  receive  only  $106.  Consequently, 
I  must  pay  a  sum,  which,  if  loaned,  will  amount  to  $106  in  a  year, 
which  is  $100,  Ans. 

An  allowance  made  by  a  creditor  to  a  debtor,  for  paying 
money  due  at  some  future  time  without  interest,  before  the 
time  Agreed  on  for  payment,  is  called  Discount,  and  the  sum 
paid  is  called  the  Present  Worth. 

2.  I  sell  a  piece  of  wild  land  in  Wisconsin,  for  $S6S,  to  be 
paid  in  4  years,  without  interest,  since  the  purchaser  is  to  re 
ceive  no  profit  from  his  purchase.  Wishing  ready  money,  1 
transfer  the  debt  to  a  third  person  for  a  sum,  which,  put  at  6 
per  cent,  interest,  for  the  time,  would  amount  to  $868;  what 
do  I  receive  for  my  debt  ? 

SOLUTION.  —  Since  $1  in  4  years  will  amount  to  $1<24,  for  every  time 
Si  '24  can  be  subtracted  from,  or  is  contained  in,  $868,  I  shall  receive 

81.     $868-j-$l<24  =  $700,  Ans.      . 


The  rule  is  evidently  the  same  as  in  the  last  paragraph. 

Questions.  —  IT  164.  What  is  the  subject  of  this  paragraph?  Repeat 
the  first  example.  Solve  it.  What  is  discount?  —  presert  worth  ?  How  is 
it  found  ?  Proof  ?  How  is  discount  feund  ? 


216  PERCENTAGE.  IF  165 

PROOF.  —  Find  the  amount  of  the  result  at  the  given  rate  and  time: 
this  amount  will  be  equal  to  the  given  sum. 

3.  How  much  ready  money  must  be  paid  for  a  note  of  $  18,  due  15 
months  hence,  discounting  at  the  rate  of  6  per  cent.? 

Ans.  $16<744+. 

4.  What  is  the  present  worth  of  $56 '20,  payable  in  1  year  8 

months,  discounting  at  6  per  cent.  1  at  4.1  per  cent.  ?  at  5 

per  cent.  ?  at  7  per  cent.  ?  at  7£  per  cent.  ?  at  9  per 

cent.?  Ans.  to  the  last,  $48'869. 

5.  What  is  the  present  worth  of  $  834,  payable  in  1  year  7  months 
and  6  days,  discounting  at  the  rate  of  7  per  cent.  ?         Ans.  $  750-J-. 

6.  What  is  the  discount  on  $  321'63,  due  4  years  hence,  discount 
ing  at  the  rate  of  6  per  cent.  ? 

NOTE.  —  If  the  present  worth  be  subtracted  from  the  given  amount,  the  re 
mainder  will  be  the  discount. 

Ans.  $62<25-f-. 

7.  Sold  goods  for  $  650,  payable  one  half  in  4  months,  and  the 
other  half  in  8  months ;  what  must  be  discounted  for  present  pay 
ment,  at  6  per  cent.  ?  Ans.  $  IS'ST^-]-- 

8.  A  merchant  purchases  in  New  York  city,  goods  to  the  amount 
of  $  5378,  on  6  months  credit,  paying  6  per  cent,  more  than  if  he  had 
paid  down.     What  would  he  have  saved  if  he  had  borrowed  money 
.it  7  per  cent,  per  annum,  with  which  to  make  his  purchase  ?     What 
would  he  save  in  20  years,  averaging  2£  such  purchases  each  year? 

Ans.  to  the  last,  $6342. 


Commission. 

5T  1G5>.  1.  A  merchant  in  Utica  receives  $988  from  a. 
house  in  New  York,  with  which  to  purchase  butter,  after  de 
ducting  for  his  services  4  per  cent,  on  what  money  he  shall 
lay  out ;  how  much  will  he  pay  for  butter  ? 

SOLUTION.  —  Of  every  $1<04  which  he  receives  he  must  lay  out  1  dol 
lar,  and  retain  4  cents,  thus  having  4  cents  for  each  dollar  which  he 
pays  for  butter.  Then,  as  many  times  as  $1<04  is  contained  in  $988,  so 
many  dollars  he  must  pay  out.  $988  -f-  $1<04  =  $950,  Ans. 

NOTE.  — Had  we  multiplied  $988  by  '04  to  ascertain  how  much  he  received 
for  his  services,  it  would  have  given  him  4  cents  on  each  dollar  which  he  re 
ceived.  This  would  have  given  him  4  cents  for  laying  out  96  cents,  instrt-Ml 
of  100  cents,  as  required  by  the  question. 

Questions.  —  ^T  165.  What  is  the  first  example  1  Give  the  solution. 
Show  the  error,  if  performed  as  supposed  in  the  note.  What  is  commission  ? 
What  are  persons,  who  buy  and  sell  goods  for  others,  called  ?  When  they 
receive  money  to  disburse,  now  is  their  commission  estimated  ?  When  the  , 
value  of  the  goods  bought  or  sold  is  known,  how  is  the  commission  esti 
mated  ? 


ff  166.  PERCENTAGE.  217 

The  per  cent,  or  amount  allowed  to  persons  for  their  ser 
vices  in  assisting  merchants  and  others  in  purchasing  and 
selling  goods,  and  for  transacting  other  business,  is  called 
Commission. 

Persons  who  buy  and  sell  goods  for  others,  are  called 
Agents,  Commission  Merchants,  Correspondents,  and  Factors. 

When  agents,  &c.,  receive  money  to  disburse,  their  com 
mission  is  estimated  in  the  same  manner  as  discount,  by 
the  rule  IT  163. 

2.  Received  $2475,  with  which  to  purchase  wool,  after  deduct 
ing  my  commission  of  5  per  cent.  ;  how  many  dollars  will  I  pay  out? 
What  will  be  the  amount  of  my  commission  ? 

Ans.  to  the  last,  $  117'857. 

3.  Sent  my  £  gent  $4820,  with  which  to  purchase  wheat,  after 
deducting  his  commission  of  7£  per  cent. ;  how  much  money  will  he 
expend,  and  what  will  be  the  amount  of  his  commission  ? 

Ans.  He  will  expend  $4483*72 -{-. 

NOTE.  —  When  the  value  of  the  goods  bought  or  sold  is  known,  the  com 
mission  is  estimated  upon  that  value,  in  the  same  manner  as  percentage. 
(IT  143.) 

4.  A  commission  merchant  sold  goods  to  the  amount  of  $  1422,  at 
5  per  cent,  commission ;  how  much  did  he  receive  for  his  services  ? 

Ans.  $71'10. 

5.  My  correspondent  sends  m'e  word  that  he  has  purchased  goods 
to  the  value  of  $  1286,  on  my  account ;  what  will  be  his  commission, 
&i  2i  per  cent.  ?  Ans.  $  32' 15. 

6.  What  must  I  allow  my  correspondent  for  selling  goods  to  the 
amount  of  $2317'46,  at  a  commission  of  3^  per  cejit.  ? 

Ans.  $75'317. 

7.  A  tax  on  a  certain  town  is  $  1627' 18,  on  which  the  collector  is 
to  receive  2£  per  cent,  for  collecting  ;  what  will  he  receive  for  col 
lecting  the  whole  tax?  Ans.  $40'679. 

The  time,  rate  per  cent.,  and  interest  being  given,  to  find 
the  principal. 

If  166.  1.  What  sum  of  money,  put  at  interest  16 
months,  will  gain  S1O50,  at  6  per  cent.  ? 

SOLUTION.  —  $1  in  16  months,  at  6  per  cent.,  will  gain  $'08;  and 
since  $'03  is  the  interest  of  $1  at  the  given  rate  and  time,  $  10<50  is  the 
interest  of  as  many  dollars  as  the  number  of  times  $<08  is  contained  in 
(can  be  subtracted  from)  $10 '50.  $10'50-j-  $>'08  =  $131<25,  the  prin 
cipal  required. 

Hence,  the 

RUL.E. 

Find  the  interest  of  $1,  at  the  given  rate  and  time,  by 
19 


218  PERCENTAGE.  H  167,  168. 

which  divide  the  givsn  interest;  the  quotient  will  be  the  prin 
cipal  required. 

EXAMPLES  FOR  PRACTICE. 

3.  A  man  paid  $4'52  interest,  at  the  rate  of  6  per  cent.,  at  the 
end  of  1  year  4  months  ;  what  was  the  principal?  Ans.  $56'50. 

3.  A  man  received,  for  interest  on  a  certain  note,  at  the  end  of  1 
year,  $20 ;  what  was  the  principal,  allowing  the  rate  to  have  been  6 
percent.?  Ans.  $333' 333  J 

4.  A  man  leases  a  farm  for  $562,  which  sum  is  10  per  cent,  of 
the  value  of  the  farm  ;  how  much  is  the  farm  worth  ? 

The  principal,  interest,  and  time  being  given,  to  find  the 
rate  per  cent. 

IF  167.     1.    If  I  pay  $3<7S  interest,  for  the  use  of  $36  for 

1  year  and  6  months,  what  is  that  per  cent.  ? 

SOLUTION.  —  The  interest  of  $36,  at  1  per  cent.,  for  1  year  and  6 
months,  is  $'54,  and  consequently  $3*78  is  as  many  per  cent,  as  the 
times  $<54  is  contained  in  $3'78.  $3;78  -j-  '54  =  7  per  cent. 

Hence,  the 

RULE. 

Find  the  interest  on  the  given  sum,  at  1  per  cent.,  for  the 
given  time,  by  which  divide  the  given  interest;  the  quotient 
will  be  the  rate  at  which  interest  is  paid. 

EXAMPLES    FOR    PRACTICE. 

2.  If  I  pay  $2'34  for  the  use  of  $468,  1  month,  what  is  the  rate 
per  cent.  '\  Ans.  6  per  cent. 

3.  At  $  46' 80  for  the  use  of  $  520, 2  years,  what  is  that  per  cent.  ? 

Ans.  4£  per  cent. 

4.  A  stockholder,  who  owned   10  shares,  of   $  100  each,  of  the 
Tonawanda  Railroad  company  stock,  received  a  dividend  of  $50  every 
6  months  ,  what  per  cent,  was  that  on  the  money  invested  ? 

Ans.  10  per  cent. 

5.  A  widow  lady,  whose  expenses  are  $324  a  year,  has  $5400 
in  money ;  at  what  rate  per  cent,  must  she  loan  it,  that  the  interest 
may  pay  her  expenses  1 

The  -principal,  rate  per  cent.,  and  interest  being  given,  to 
find  the  time. 

IF  168.  1.  The  interest  on  a  note  of  $36,  at  7  per  cent., 
was  $3'78 ;  what  was  the  time  ? 

Questions.  —  TI  166.  When  the  time,  rate  per  cent.,  and  interest  are 
given,  how  may  the  principal  be  found  ?  Explain  the  principles  of  the  rule. 

IT  167.  When  the  principal,  interest,  and  time  are  given,  how  do  you  find 
Ihe  rate  per  cent.  ?  Explain  the  principles  of  the  rule. 


T  169.  PERCENTAGE.  219 

SOLUTION.  —  The  interest  of  $36,  1  year,  at  7  per  cent.,  is  $2(52. 
Since  $2<52  will  pay  for  the  use  of  $36  1  year,  S3'78  will  pay  for  the  use 
cf  it  as  many  years  as  the  times  $2'52  is  contained  in  $3<78.     $3<78-s 
2<52  _:l</>  years,  =  1  year  6  months,  the  time  required. 

Hence,  the 

RUL.E. 

Find  the  interest  for  1  year  on  the  principal  given,  at  the 
given  rate,  bx  which  divide  the  given  interest  ;  the  quotient 
will  be  the  time  required,  in  years  and  decimal  parts  of  a 
year. 

EXAMPLES    FOR    PRACTICE. 

2.  If  $  31<71  interest  be  paid  on  a  note  of  $226-50,  what  was  the 
time,  the  rate  being  6  per  cent.? 

Ans.  2<33£  =  2  years  4  months. 

3.  On  a  note  of  $600,  paid  interest  $20,  at  8  per  cent.  ;  what 
was  the  time  1 

Ans.  '416  -j-  yr.  =  5  mo.,  nearly.  It  would  be  exactly  5  but  for 
the  fraction  lost. 

4.  The  interest  on  a  note  of  $  217'25,  at  4  per  cent.  ,  was  $28'242  ; 
what  was  the  time?  Ans.  3  years  3  months. 

NOTE.  —  When  the  rate  is  6  per  cent.,  we  may  divide  the  interest  by  4  the 
principal,  and  the  quotient^  removing  the  separatrix  two  places  to  the  left, 
will  be  the  answer  required,  in  months  and  decimals  of  a  month. 

The  percentage  of  any  number  of  dollars  being  given,  to 
find  the  rate. 

^F  16O.  1.  A  merchant  purchases  a  piece  of  broadcioth 
for  $60  ;  what  will  be  the  per  cent,  of  gain,  if  he  sells  it  for 
867*20  ? 

$67<20 
GO'OO 

-  SOLUTION.  —  Subtracting  the  price  which  he  gave 

60'  )  7'20  (  '12     from  the  Price  for  which  he  sells  the  cloth,  we  have 

' 


60  '  a  on  ®60>  °f  wmctl  we  must  take 

•fa  for  the  gain  on  Si.  "We  get  12  cents  as  the  gain 
on  $1,  or  100  cents.  Hence  the  gain  is  12  hundredths 
of  the  sum  paid,  or,  Ans.  12  per  cent. 

120  Hence,  the     ' 

0 

RULE. 

Divide  the  percentage  of  the  number  of  dollars  by  the  num- 

Questions.  —  IT  168.    When  the  principal,  rate  per  cent.,  and  interest 
are  given,  how  do  you  find  tb«  time  ?    Explain  the  principles  of  the  rule. 


220  PERCENTAGE.  IF  170 

her  of  dollars  on  which  it  has  accrued ;  the  quotient,  which  is 
the  percentage  of  $1,  or  100  cents,  will  express  the  rate  per 
cent. 

EXAMPLES    FOR    PRACTICE. 

2.  A  merchant  purchases  goods  to  the  amount  of  $  550  ;  what  per 
cent,  profit  must  he  make  to  gain  $66?  Ans.  12  per  cent. 

3.   What  per  cent,  profit  must  he  make  on  the  same  purchase, 

to  gain  $38'50?  to  gain  $24'75?  to  gain  $2'75? 

Ans.  to  the  last,  '005,  or  &  per  cent. 

4.  Bought  a  hogshead  of  rum,  containing  114  gallons,  at  96  cents 
per  gallon,  and  sold  it  again  at  $  T0032  per  gallon ;  what  was  the 
whole  gain,  and  what  was  the  gain  per  cent.  ? 

A        {  $4'924,  the  whole  gain. 
J>  \        4£,  gain  per  cent. 

5.  Bought  30  hogsheads  of  molasses,  for  $  600  ;  paid  in  duties 
$20'66;  for  freight,  $40'78;  for  porterage,  $6'05,  and  for  insur 
ance,  $  30' 84  ;  if  I  sell  them  at  $  26  per  hogshead,  how  much  shall 
I  gain  per  cent.  ?  Ans.  11'695  -j-  per  cent. 

6.  A  spendthrift,  who  received  an  inheritance  of  $3000,  spent 
$  960  the  first  week  in  gambling ;  what  per  cent,  of  his  money  is 
gone  ?  Ans.  32  per  cent. 

7.  A  farmer  paid  $  2'50  for  insuring  buildings  worth    $  1000 ; 
what  was  the  rate  per  cent.  ?  Ans.  £  per  cent. 

8.  A  commission  merchant  receives  $  37'50  for  selling  goods  to 
the  amount  of  $  1250  ;  what  was  the  rate  per  cent.  ? 

Ans.  3  per  cent. 

9.  A  broker  receives  $270  for  selling  $18000  worth  of  stocks; 
what  is  the  per  cent,  for  brokerage  ?  Ans.  l£  per  cent. 


Bankruptcy. 

^T  17O.  An  individual  who  fails  in  business  sometimes 
makes  an  assignment  of  his  property,  which  is  divided  among 
his  creditors  according  to  their  respective  debts. 

In  making  calculations  in  bankruptcy,  we  find  what  can  be 
paid  on  each  dollar  owed,  and  multiply  this  by  the  number  of 
dollars  which  each  man  claims,  to  get  his  share. 

EXAMPLES  FOR  PRACTICE. 

1.  An  extensive  banking  house  in  New  York  fails  for  $800,000  , 
the  property  of  the  house  is  found  to  be  $  300,000  ;  what  is  paid  on 
a  dollar?    '  Ans.  $'37£,  or  37£  per  cent. 

2.  How  much  will  a  man  receive  on  the  above,  whose  dues  aro 
$16500?  Ans.  $6187'50. 

Questions.  -    V  169.    How  is  Ex.  1  explained  ?    Rule. 


1l  171.  PERCENTAGE.  221 

3.  A  merchant  fails  in  business,  owing  to  A,  $  250 ;  to  B,  $  320  ; 
M  C,  $  500  ,  to  D,  $  180  ;  to  E,  $  700  ;  to  F,  $  390  ;  to  G,  $  65-50 ; 
jo  H,  $  1300  ;  to  I,  $2200  ;  to  J,  $850 ;  his  property  is  found  to 
be  $  4653  ;  how  much  does  each  receive  ? 


General  Average. 

IT  171  •  When  a  ship  is  in  distress,  the  expense  incurred, 
or  the  damage  suffered  by  the  ship,  or  any  part  of  tine  cargo, 
is  averaged  upon  the  value  of  the  ship  ;  upon  the  cargo,  esti 
mated  at  what  the  goods  will  bring  at  the  destined  port ;  and 
upon  the  freight,  deducting  one  third,  on  account  of  the  sea 
men's  wages.  • 

To  estimate  general  average,  we  find  the  proportion  of  the 
loss  on  each  dollar,  and  then  the  loss  on  the  number  of  dol 
lars  of  each  contributary  interest. 

The  ship  Silas  Richards,  in  her  voyage  from  New  York  to  Charles 
ton,  became  stranded  on  the  coast  of  North  Carolina,  when  it  was 
found  necessary  to  throw  overboard  506  barrels  of  flour,  belonging  to 
Goodrich  &  Co.,  worth  $6'87  per  barrel.  The  expense  of  getting 
the  vessel  off,  was  $  197  ;  of  supplying  new  rigging,  $240,  of  which 
one  third  is  to  be  deducted,  as  the  new  is  supposed  to  be  better  than 
the  old.  The  ship  is  worth  $10232;  the  freight  is  $4800,  of 
which  one  third  is  to  be  deducted.  Goodrich  &  Co.  had  on  board 
1000  barrels  of  flour;  M.  H.  Newman  &  Co.,  goods  worth  $4000  ; 
D.  Appleton,  goods' worth  $5236;  Hyde  &  Duren,  goods  worth 
$  9000  ;  how  much  do  Goodrich  &  Co.  realize  for  all  their  flour ; 
what  does  each  interest  contribute  towards  the  loss,  and  what  is  the 
rate  per  cent,  on  the  contributary  interests? 

Goodrich  &  Co.  realize  $6186'6G9. 

The  ship's  portion  of  the  loss  is        $1017'735-|- 
Portiori  of  the  freight,  $318'291  -  - 

"         M.  H.  Newman  &  Co.,     $397*863-- 
"         D.  Appleton,  $520'803-- 

"         Hyde  &  Duren,  $895' 193 -- 

"         Goodrich  &  Co.,  $683'33l  + 

Rate  per  cent.,  9 Iff a- 

Questions*  —  IT  17O.  What  do  you  understand  by  bankruptcy  ?  How 
are  calculations  in  bankruptcy  made? 

IT  171.  What  do  you  understand  by  general  average?  What  expenses 
and  losses  may  it  embiace  ?  On  what  r '  Afferent  interests  are  tae  expenses 
averaged  ? 

19* 


222  PERCENTAGE.  V  172. 


Partnership. 

IF  179.  When  two  or  more  persons  unite  a  part  or  the 
whole,  of  their  capital  for  the  prosecution  of  business,  they 
form  a  Company  or  Firm,  and  their  business  is  called  Partner 
ship  business.  Each  member  of  a  firm  is  called  a  partner. 

Capital  or  Stock  is  the  money  or  other  property  employed 
in  trade;  Joint  Stock  is  stock  of  a  company  or  firm.  Divi 
dend  is  the  gain  or  loss  to  be  shared  among  the  partners. 

1.  Two  persons  have  a  joint   stock  in  trade;  A  put  in 
$250,  and  B  $350 ;  they  gain  $150 ;  what  is  each  man's 
share  ? 

OPERATION.  SOLUTION.  —  We  di- 

6iOO)il!50'00  vide   the   whole   &ain' 

$150,  by  600,  which  will 

give  us  the  gain  on  $1, 

$'25  =$'25.  Then  250  times 

A's  gain,  -25  X  250  ==  $62'50,  )     ,  $'25  is  A's  gain,  and  350 

£'A  gain,  '25  X  350  =  $87'50,  J  Ans'     times  $'25  is  B's  ?ain- 

By  the  first  operation, 

we  get  the  rate  per  cent,  of  gain  or  loss,  according  to  If  169.     The  sec 
ond  may  be  performed  by  the  ordinary  rule  for  percentage. 
Or  the  operation  may  be  performed  as  follows  : 

A's  gain  will  be  f  £$  =  -&  of  $150  =  $62'50. 

Rs  gain  will  be  f  ££  =  -fa  of  $150  =  $87'50.      Hence, 

RULE. 

Take  such  a  part  of  the  whole  gain  or  loss  as  each  man's 
stock  is  part  of  the  whole  stock ;  the  part  thus  taken  will  be 
his  share  of  the  gain  or  loss. 

NOTE.  — This  rule  may  be  applied  to  tne  operations  in  several  preceding 
paragraphs. 

EXAMPLES    FOR    PRACTICE. 

2.  A,  B,  and  C  trade  in  company ;  A's  capital  was  $  175,  B's 
$200,  and  C's  $500  ;  by  misfortune  they  lose  250  ;  what  loss  must 
each  sustain?  ($50,  A's  loss. 

Ans.{  $  57'142f,    B's  loss. 
(  $142'857|,    C's  loss. 

3.  Divide  $600  among  3  persons,  so  that  their  shares  may  be  to 
each  other  as  1,  2,  3,  respectively.     Ans.  $  100,  $200,  and  $300. 

Questions.  —  IT  172.  What  is  a  company  or  firm'/  —  partnership  busi 
ness?  —  a  partner?  —  capital  or  stocL?  — joint  stock?  — dividend?  Give 
the  first  solution  of  Ex.  1  ;  —  the  seo  .nd.  Rule. 


f  173.  PERCENTAGE.  223 

4.  Two  merchants,  A  and  B,  loaded  a  ship  with  500  hhds.  of 
rum;  A  loaded  350  hhds.,  and  B  the  rest;  in  a  storm,  the  seamen 
were  obliged  to  throw  overboard  100  hhds. ;  how  much  must  each 
sustain  of  the  loss?  Ans.  A  70,  and  B  30  hhds. 

5.  A  and  B  companied  ;  A  put  in  $45,  and  took  out  f-  of  the 
gain  ;  how  much  did  B  put  in?  Ans.  $ 30. 

NOTE.  —  They  took  out  in  the  same  proportion  as  they  put  in  ;  if  3  fifths 
of  the  stock  is  $45,  how  much  is  2  fifths  of  it? 

6.  A  and  B  companied,  and  traded  with  a  joint  capital  of  $  400 ;  A 
received,  for  his  share  of  the  gain,  £  as  much  as  B  ;  what  was  the 
stock  of  each?       <?  !  /  •  A    c  5  $  133'333£,  A's  stock. 

Ans-  I  $266<666|,  B's  stock. 

7.  A  and  B  ventured  equal  stocks  in  trade,  and  cleared  $  164 ; 
by  agreement,  A,  because  he  managed  the  concerns,  was  to  have  $5 
of  the  profits,  as  often  as  B  had  $  2  ;  what  was  each  one's  gain?  and 
how  much  did  A  receive  for  his  trouble  ? 

Ans.  A's  gain  was  $117<142f,  and  B's  $46'857|,  and  A  re 
ceived  $70'285f-  for  his  trouble. 

8.  A  cotton  factory,  valued  at  $  12000,  is  divided  into  100  shares; 
if  the  profits  amount  to  15  per  cent,  yearly,  what  will  be  the  profit 

accruing  to  1  share  ?  to  2  shares  ?  to  5  shares  ?  to  25 

shares?  Ans.  to  the  last,  $450. 

9.  In  the  above-mentioned  factory,  repairs  are  to  be  made  which 
will  cost  $340;  what  will  be  the  tax  on  each  share,  necessary  to 

raise  the  sum?   on  2  shares?    on  3  shares?   on  10 

shares?  Ans.  to  the  last,  $34. 

10.  Two  men  paid  10  dollars  for  the  use  of  a  pasture  1  month  ;  A 
kept  in  24  cows,  and  B  16  cows ;  how  much  should  each  pay? 

PARTNERSHIP  ON  TIME. 

IT  173.     1.    Two  men  hired  a  pasture  for  $10  ;  A  put  in 

8  cows  3  months,  and  B  put  in  4  cows  4  months  ;  how  much 
should  each  pay  ? 

SOLUTION.  —  The  pasturage  of  8  cows  for  3  months  is  the  same  as  of 
24  cows  for  1  month,  and  the  pasturage  of  4  cows  for  4  months  is  the 
same  as  of  16  cows  for  1  month.  The  shares  of  A  and  B.  therefore,  are 
24  to  16,  as  in  the  former  question.  Hence, 

When  time  is  regarded  in  partnership,  multiply  each  one's  stock  by  the 
time  he  continues  it  in  trade,  and  use  the  product  for  his  share. 

Ans.  A  6  dollars,  and  B  4  dollars. 

2.  A  and  B  enter  into  partnership ;  A  puts  in  $  100  6  months,  and 
then  puts  in  $  50  more ;  B  puts  in  $  200  4  months,  and  then  takes 
out  $  80  ;  at  the  close  of  the  year  they  find  that  they  have  gained 
$  95  ;  what  is  the  profit  of  each  ?  A  <  $  43'71 1,  A's  share. 

'•  \  $51'288,  B's  share. 

Questions.  —  IT  173.  What  are  we  to  understand  by  partnership  05 
time  ?  How  do  we  proceed  ? 


224  PERCENTAGE.  1F 174. 

3.  A,  with  a  capital  of  $500,  began  trade  Jan.  1,  1846,  and, 
meeting  with  success,  took  in  B  as  a  partner,  with  a  capital  of  $  600, 
on  the  first  of  March  following ;  four  months  after,  they  admit  C  aa 
a  partner,  who  brought  $  800  stock  ;  at  the  close  of  the  year,  the* 
find  the  gain  to  be  $  700  ;  how  must  it  be  divided  among  the  part 
ners*  (  $250,  A "s  share. 

Ans.  \  $250,  B's  share. 
(  $200,  C's  share. 


Banking. 

IT  174.  A  bank  is  an  incorporated  institution  which  traf» 
fics  in  money.  Bank  notes,  or  bank  bills  are  promissory 
notes,  payable  to  bearer. 

Banks  loan  their  money  on  notes,  the  interest  always  being 
paid  in  advance. 

For  example,  B  holds  A's  note  for  $100,  payable  in  90  days, 
without  interest.  But  B  is  in  immediate  want  of  money. 
He  carries  A's  note  to  a  bank,  and  if  their  credit  be  un 
doubted,  the  bank  will  receive  A's  note  and  pay  the  face  of  it, 
minus  the  interest  on  it,  for  3  days  more  than  the  given  time, 
(90  -[-  3  =  93  days.)  The  note  is  then  said  to  be  discounted 
at  the  bank.  These  3  days  are  called  days  of  grace. 

But  the  bank  will  require  B  to  write  his  name  on  the  back 
of  the  note.  This  is  called  endorsing  the  note.  It  subjects 
B  to  pay  the  note  when  the  90  days  and  the  3  days  of  grace 
shall  expire,  provided  A,  who  gave  the  note,  should  fail  so 
to  do. 

The  money  received  from  the  bank  for  the  note,  is  called 
the  avails  of  the  note.  The  note  is  said  to  be  mature  when 
the  time  that  it  should  be  paid  shall  arrive. 

Again  :  B  as  principal,  with  C  and  D  as  sureties,  may  give 
their  note  jointly  and  severally  to  the  bank,  for  the  sum 
wanted,  payable  at  a  specified  time,  without  interest.  Then 
if  B  fails  to  pay  the  note,  his  sureties,  C  and  D,  either  or  both, 
will  be  holden  to  pay  it. 

SIOO'OO  Principal.  NOTE.  -  Bank  interest, 

when  the  rate  is  6  per 

$1<00  Int.  60  days.  cent->  may  be,  past  by  in- 

<^f)      "      «*0  Arw  snection,  as  follows  :  Let 

ou  aab.  sum  on  which  it  is  to 


'05     "      3  days,  of  grace.  t«  cast,  be  $100.     The 

principal  itself,  removing 


$1'55  Discount  for  90  days  and  grace,     the  point  two  places  to  the 

left,  is  made  to  express  the 
interest  for  60  days,  (Sl'OO  )  half  of  which  ($'50)  is  the  interest  for  30  day% 


1T  175.  PERCENTAGE.  225 

and  T^of  the  interest  for  30  days,  $'50  -r  10  =$'05,  is  the  interest  for  3 
da  vs. 

The  sum  of  these  results  is  the  bank  interest  on  $100,  at  6  per  cent.,  for  90 
da.ys,  which  sum  $1'55,  deducted  from  the  face  of  the  note,  makes  its  avails  to 

" 


,. 

If  the  discount  he  other  than  6  per  cent.,  take  such  fractional  part  of  the  dis 
count  a*  *  per  cent,  as  the  required  rate  is  less  or  more  than  6  per  cent.,  which 
added  to  or  subtracted  from  the  discount  at  G  per  cent.,  as  the  case  may  re 
quire,  will  give  the  discount  sought. 

EXAMPLES    FOR    PRACTICE. 

1.  What  will  be  received  on  a  bank  note  of  $  500,  due  in  90  days, 
at  7  per  cent.?  .  Ans.  $490*55$. 

2.  What  is  the  discount  of  a  bank  note  for  $300,  due  in  90  days, 
at  5  per  cent.  ?  Ans.  $  3'875. 

3.  What  is  the  discount  of  a  note  for  $600,  due  in  90  days,  at  8 
percent.?  Ans.  $  12'40. 

4.  What  is  received  on  a  note  for  $  740,  due  in  90  days,  at  6  per 
cent.?  Ans.   $728'53. 

5.  A  man  gets  a  note  of  $  1000  discounted  for  90  days,  at  6  per 
cent,  per  annum,  and  lends  the  money  immediately,  till  the  time 
when  he  is  obliged  to  pay  ;  what  does  he  lose?         Ans.  $  '24  -{-. 


Taxes. 

5T  175.  A  tax  is  a  sum  imposed  on  an  individual  for  a 
public  purpose. 

A  Poll  tax  is  a  specific  sum  assessed  on  male  citizens  above 
21  years  of  age ;  each  person  so  assessed  is  called  a  poll. 

Taxes  are  usually  assessed  either  on  the  person  or  prop 
erty  of  the  citizens,  and  sometimes  on  both. 

Property  is  of  two  kinds,  personal  property  and  real 
estate. 

Personal  is  movable  property,  such  as  money,  notes,  cattle, 
furniture,  &c. 

Real  estate  is  immovable  property,  such  as  lands,  houses, 
stores,  &c. 

An  Inventory  is  a  list  of  articles. 

In  assessing  taxes,  it  is  necessary  to  have  an  inventory  of 
all  the  taxable  property,  both  personal  and  real,  of  those  on 
whom  the  tax  is  to  be  levied,  and  also,  (if  a  poll  tax  is  to  be 
raised,)  of  the  whole  number  of  polls ;  arid  as  the  polls  are 

Questions.  —  IT  174.  What  is  a  bank?  When  is  bank  interest  paid? 
Illustrate  by  the  example.  What  is  meant  by  days  of  grace  ?  How  long  is 
the  interest  cast  on  a  note  due  in  90  days  ?  What  security  does  the  bank  re 
quire  ?  How  is  bank  interest  found  by  inspection,  when  the  rate  is  6  per 
cent.  ?  How,  when  it  is  any  other  rate  7 


226 


PERCENTAGE. 


1T175. 


rated  as  a  certain  sum  each,  we  must  first  deduct  from  the 
whole  tax  the  amount  of  the  poll  tax,  and  the  remainder  is  to 
be  assessed  on  the  property. 

The  tax  on  $1  is  found  by  dividing-  the  amount  to  be  as 
sessed  on  the  property  by  the  value  of  the  property  taxed. 

The  tax  on  any  amount  of  property  is  found  by  multiplying 
the  value  of  the  property  by  the  tax  on  $1. 

1.  A  tax  of  $917  is  to  be  assessed  on  a  town  in  which  are 
320  polls,  assessed  at  40  cents  each ;  the  inventoried  value 
of  the  personal  and  real  property  of  the  town  is  $52600 ; 
what  is  the  amount  of  the  poll  taxes  ?     How  much  remains  to 
be  assessed  on  the  property?     What  is  the  tax  on  $1  ? 

SOLUTION.  $'40  X  320  =  $128,  amount  of  poll  tax;  then  $917  — $128 
=  $789,  amount  to  be  assessed  on  property,  and  $789  -5- $52600  = 
$'015,  the  tax  on  $1. 

NOTE.  —  In  making  out  a  tax  list,  form  a  table  containing  the  taxes  on  1 , 2, 
3.  &c.,  to  10  dollars:  then  on  20,  30,  &c.,  to  100  dollars;  and  then  on  100, 
200,  &c.,  to  1000  dollars.  Then,  knowing  the  inventory  of  any  individual,  it 
is  easy  to  find  the  tax  upon  his  property. 

Let  us  apply  this  method  in  assessing  a  tax  on  a  town. 

2.  A  certain  town,   valued  at  $64530,   raises  a  tax  of 
$2259'90 ;  there  are  540  polls,  which  are  taxed  $'60  each ; 
what  is  the  tax  on  a  dollar,  and  what  will  be  A's  tax,  whose 
real  estate  is  valued  at  $1340,  his  personal  property  at  $874, 
and  who  pays  for  2  polls  ? 

SOLUTION.  540  X  $  '60  =  $324,  amount  of  the  poll  taxes,  and 
$2259'90  — $324  =  1935'90,  to  be  assessed  on  property,  and  $1935'90 
-i- $64530  =$'03,  tax  on  $1. 

TABLE. 


dolls. 

dolls. 

dolls. 

dolls. 

dolls. 

dolls. 

Tax  on  i  i 

s  '03 

Tax  on  10  i 

s    '30 

Tax 

on  100  i 

s    3' 

«      2 

'06 

«      20 

'60 

ft 

200 

<    6< 

«      3 

'09 

«      30 

'90 

u 

300 

'    9' 

«      4 

'12 

«      40 

1'20 

a 

400 

'  12' 

«      5 

'15 

«      50 

1-50 

u 

500 

'  15' 

«      6 

'18 

«      60 

1'80 

(( 

600 

'  18' 

«      7 

'21 

«      70 

2'10 

K 

700 

'  21' 

«      8 

'24 

«      80 

2  '40 

it 

800 

'  24' 

«      9 

'27 

«      90 

2'70 

tc 

900 

'  27' 

" 

1000 

'  30' 

Questions.  —  IT  175.  What  is  a  tax  ?  —  a  poll  tax  ?  How  are  taxes 
usually  assessed  ?  '  Property  is  of  what  kinds  ?  What  is  personal  property  ? 
—  real  estate  ?  What  is  au  inventory  ?  In  assessing  taxes,  what  is  to  be 
done  ?  When  a  poll  tax  is  to  be  raised,  what  must  first  be  done  ?  How  do 
you  find  the  amount  to  be  assessed  on  the  property  ?  How  do  you  find  the  tax 
on  $1 '?  How  on  any  amount  of  property  ?  What  course  is  usually  pursued 
by  assessors  in  making  out  a  lax  list  ?  Explain  the  manner  in  which  any  in 
dividual's  tax  is  made  out  from  the  assessor's  tax  tab.e.  How  may  a  tax  list 
be  proved? 


IT  176.  PERCENTAGE.  227 

Now  to  find  A's  tax,  his  real  estate  t«ing  $1340;  we  find,  by  the 
table,  that 

The  tax  on        .        .      $1000  is  $30< 

The  tax  on        .        .  300        .        .         "  9'' 

The  tax  on  40        .        .        «  ,  1'20 


Tax  on  his  real  estate, $40 '20 

In  like  manner,  we  find  the  tax  on  his  personal  property 

to  be 26<22 

2  polls  at '60  each,  are    ....         .         . 

Amount,  $67<62 

3.  What  will  be  the  amount  of  B's  tax,  of  the  same  town,  whose 
inventory  is  874  dollars  raz/,  and  210  dollars  personal  property,  and 
who  pays  for  3  poll's  ?  Ans.  $34' 32. 

4.    of  C's  paying  for  2  polls,  whose  property  is  valued  at 

$3482?  — •• —  of  D's,  paying  for  1  poll,  whose  pioperty  is  valued  at 
$4675?  Ans.  to  the  last,  $140'85. 

PROOF.  —  After  a  tax  list  is  made  out,  add  together  the  taxes  of  all  the 
individuals  assessed ;  if  the  amount  is  equal  to  the  whole  tax  assessed 
the  work  is  right. 


Duties. 

IT  176.     Duties  or  customs  are  taxes  on  imported  goods. 

A  custom-house  is  a  house  or  an  office  where  customs  are 
paid. 

Government  has  established  a  custom-house  at  every  port 
in  the  United  States  into  which  foreign  goods  are  imported. 

Besides  duties  on  merchandise,  every  vessel  employed  in 
commerce  is  required  to  pay  a  certain  sum  for  entering  the 
ports.  This  sum  is  in  proportion  to  the  size  or  tonnage  of 
the  vessel. 

The  income  to  the  government,  from  duties  and  tonnage,  it, 
called  Revenue. 

All  duties  are  imposed  and  regulated  by  the  general  gov 
ernment,  and  must  be  the  same  in  all  parts  of  the  Union. 

NOTE.  —  A  table  of  the  duties  imposed  by  government  is  called  a  Tariff. 
The  law  requires  that  the  cargoes  of  all  vessels  engaged  *n 


Questions.  —  IT  176.  What  are  duties  ?  —  custom-houses,  where,  and 
by  whom  established?  What  tax  is  named  besides  duties,  and  how  propor 
tioned  ?  What  is  revenue  ?  How  are  duties  imposed  ?  What  is  a  tariff? 
How  is  the  value  of  the  ^oods  in  a  vessel  ascertained  ?  How  many  kinds  of 
duties?  What  are  specific  duties?  — ad  valorem  duties?  Define  ad  valo 
rem. 


228  PERCENTAGE.  1[  177. 

foreign  commerce,  shall  be  weighed,  measured,  or  gauged,  by 
the  custom-house  officers,  for  the  purpose  of  ascertaining  the 
amount  or  value  of  the  goods  on  which  duties  are  to  be  paid. 

Duties  on  imported  goods  are  of  two  kinds,  Specific  and 
Ad  Valorem. 

A  Specific  duty  is  a  certain  sum  per  ton,  hundred  weight> 
pound,  hogshead,  gallon,  square  yard,  foot,  &c.,  without  re 
gard  to  its  value. 

Ad  Valorem  signifies  upon  the  value. 

An  Ad  Valorem  duty  is  a  certain  per  cent,  on  the  sum  paid 
for  the  goods  in  the  country  from  which  they  are  brought. 

SPECIFIC  DUTIES. 

IT  ITT1.    In  the  custom-house  weight  and  gauge  of  goods, 
certain  deductions  are  made  for  the  box,  bag,  cask,  &c.,  con 
taining  the  goods,  and  also  for  leakage,  breakage,  &c.    These 
deductions  must  be  made  before  the  specific  duties  are  im 
posed. 

Gross  weight  is  the  weight  of  the  goods  together  with  the 
box,  bale,  bag,  cask,  &c.,  which  contains  them. 

Draft  is  an  allowance  made  for  waste,  which  is  to  be  de 
ducted  from  the  gross  weight,  and  is  as  follows : 

On        112  Ibs.  1  lb. 

Above  112  Ibs.,  and  not  exceeding  224  Ibs.,  2  Ibs. 

"       224  Ibs.,       "  "  336  Ibs.,  3  Ibs. 

"       336  Ibs.,       "  «         1120  Ibs.,  4  Ibs. 

"     1120  Ibs.,       "  "         2016  Ibs.,  7^1bs. 

"    2016  Ibs.  9  Ibs. 

Tare  is  an  allowance  for  the  weight  of  the  box,  bag,  cask, 
&c.  It  is  to  be  deducted  from  the  remainder  of  any  weight 
or  measure,  after  the  draft  or  tret  has  been  allowed. 

Leakage  is  an  allowance  of  2  per  cent,  on  all  liquors  in 
casks,  paying  duty  by  the  gallon. 

Breakage  is  an  allowance  of  10  per  cent,  on  ale,  beer,  and 
porter  in  bottles,  and  of  5  per  cent,  on  all  other  liquors  in  bot 
tles  ;  or  the  importer  may  have  the  bottles  counted,  and  pay 
duties  on  the  number  remaining  unbroken. 

Questions.  —  IT  177»  What  deductions  are  made,  if  goods  pay  specific 
duties  ?  What  is  RTOSS  weight  ?  —  draft,  or  tret  1  How  much  is  it,  and 
when  deducted?  What  is  tare,  and  when  deducted?  What  is  leakage?  — 
breakage,  and  wnat  privihige  has  the  importer  ?  —  net  weight  ?  Rule  for  cal 
culating  specific  duties. 


t  178.  PERCENTAGE.  229 

Net  weight  is  the  weight  of  the  goods  after  deducting  the 
weight  of  the  box,  bale,  &c.,  and  making  all  other  allow 
ances. 

1.  What  is  the  specific  duty  on  5  hogsheads  of  molasses, 
each  containing  120  gallons,  at  12J  cents  per  gallon,  the  cus 
tomary  allowance  being  made  for  leakage  ? 

SOLUTION.  — Since  there  are  120  gallons  in  1  hogshead,  in  5  hogs 
heads  there  are  5  times  120  gallons  =  600  gallons.  2  per  cent,  of  600 
gallons  is  600  X  (°2  =  12  gallons,  and  600  gallons—  12  gallons  =  588 
gallons.  Since  the  duty  on  1  gallon  is  12J  cents,  the  duty  on  588  gal 
lons  is  588  times  $'125  =  $73<50.  Hence, 

To  find  the  specific  duty  on  any  given  quantity  of  goods, 

RTJL.E. 

1.  Deduct  from  the  given  quantity  of  goods  the  legal  a.- 
lowance  for  draft,  tare,  leakage  or  breakage. 

II.  Multiply  the  remainder  by  the  duty  on  a  unit  of  the 
weight  or  measure  of  the  goods,  and  the  product  will  be  the 
duty  required. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  specific  duty  on  75  barrels  of  figs,  each  weighing 
83  pounds    gross,  tare  in  the   whole  597   pounds,   at  4  cents  per 
pound?  Ans.  $225' 12. 

3.  What  is  the  duty  on  420  dozen  bottles  of  porter,  at  5£  cents  per 
bottle,  the  customary  allowance  being  made  for  breakage  ? 

Ans.  $249<48. 

4.  What  is  the  duty  on  8  hogsheads  of  sugar,  each  weighing  10 
cwt.  2  qrs.  gross,  tare  14  Ibs.  per  cwt.,  at  2£  cents  per  pound? 

Ans.  $185'22. 

5.  What  is  the  duty  on  4  barrels  of  Spanish  tobacco,   the  first 
weighing  171  pounds  gross,  the  second  125  pounds  gross,  the  third 
109  pounds  gross,  and  the  fourth  99  pounds  gross,  at  6£  cents  per 
pound,  the  customary  allowance  being  made  for  draft,  and  16  pounds 
per  barrel  for  tare  ?  Ans.  $27*25. 

AD  VALOREM  DUTIES. 

U  178.  Since  ad  valorem  duties  are  estimated  upon  the 
ictual  cost  of  the  goods,  it  is  plain  that  they  are  found  by 
simply  multiplying  the  cost  of  the  goods  by  the  given  per 
cent. 

Questions.  —  IT  178.  Upon  what  are  ad  valorem  duties  estimated  ) 
How  are  they  found  ?  In  estimating  ad  valorem  duties,  are  any  allow 
ances  eVer  made  ? 

20 


230  PERCENTAGE.  IT  179. 

NOTE.  —  In  estimating  ad  valorem  duties,  no  deduct  ons  of  any  Kind  aie  to 
be  made. 

1.    What  is  the  ad  valorem  duty,  at  25  pe  r  cent ,  on  32  yds. 
of  English  broadcloth,  which  cost  $4'75  pe:  yard? 

SOLUTION.  —  32  yds.  a!  $4<75,  cost  4<75  X  32  =  1 152,  and  25  per  cent, 
of  152,  is  $38,  Ans. 

EXAMPLES  FOR  PRACTICE. 

2.  What  is  the  ad  valorem  duty,  at  18  per  cent.,  on  4(£  bags  of 
Javn  coffee,  each  weighing  115  pounds,  and  which  cost  11£  cents  per 
poui!«l  ?  Ans.  $93' 15. 

3.  What  is  the  ad  valorem  duty,  at  33&  per  cent.,  on  i' gross  of 
Sheffield  cutlery,  which  cost  $256*80  1  An5/$85'60. 

4.  What  is  the  duty,  at  20  per  cent.,  on  a  piece  of  Turkey  carpet 
ing,  containing  140  yards,  and  which  cost  $1'92  per  yard?     What 
is  the  duty  on  1  yard?     For  how  much  must  it  be  sold  per  yard,  to 
gain  25  -per  cent,  on  the  cost  and  duty  ?       Ans,  to  the  last,  $-2*88. 

5.  What  is  the  duty,  at  35  per  cent.,  on  a  case  of  Italian  silks, 
which  cosi  $4821'50?  Ans.  $1687'52£. 

6.  Wh:it  is  the  duty,  at  15  per  cent.,  on  3  dozen  gold  watches, 
which  cost  $68  each?  Ans.  $367'20. 

7.  What  Is  the  duty,  at  22  per  cent.,  on  75  chests  of  tea,  the  net 
weight  of  each  chest  being  92  pounds,  and  the  tea  costing  41  cents 
per  pound?  Ans.  $622' 38 


IT  179.    Review  of  Percentage. 

Questions.  — What  is  meant  by  percentage  ?  —  rate  per  cent.  ?  — 
general  rule?  What  are  calculated  by  percentage?  What  is  insurance? 
—  mutual  insurance?  What  is  meant  by  stocks?  —  brokerage?  — 
profit  and  loss  ?  What  is  interest,  and  how  calculated  ?  How  is  6  per 
cent,  interest  calculated  by  inspection  ?  What  is  meant  by  partial  pay 
ments  ?  What  difference  between  the  U.  S.  and  Conn,  rules  ?  To  what 
case  does  the  Vt.  rule  for  partial  payments  apply  ?  What  is  done  when 
notes  with  partial  payments  are  paid  within  a  year  ?  Ho\v  does  com 
pound  differ  from  annual  interest  ?  Like  Avhat  case  in  interest  are  dis 
count  and  commission  ?  What  do  you  say  of  bankruptcy  ?  —  of  general 
average?  —  partnership?  How  does  the  calculation  of  bank  interest 
differ  from  that  of  other  interest  ?  How  are  taxes  assessed  ?  How  do 
duties  differ  from  other  taxes?  What  two  krjds  of  duties,  and  how  is 
each  computed  ? 

EXERCISES. 

1.  'What  is  the  interest  of  $273'51,  for      year  and  10  days,  ax  7 
per  cent.  ?  Ans.  $19'677-j- 

2.  What  is  the  interest  of  $486,  for  1  year  3  months  19  days,  at  8 
percent.?  Ans.  $50'652. 


1!  179.  PERCENTAGE.  231 


3.  What  is  the  interest  of  $1600,  for  1  year  and  3  months,  at  6 
per  cent.  ?  Ans.  $120. 

4.  What  is  the  interest  of  $5'811,  for  1  yeai  11  months,  at  6  per 
cent.1?  Ans.  $'668. 

5.  What  is  the  interest  of  $2'29,  for  1  month  19  days,  at  3  per 
cent.1?  Ans.  '009. 

6.  What  is  the  interest  of  $18,  for  2  years  14  days,  at  7  per  cent.  * 

Ans.  $2'569. 

7.  What  is  the  interest  of  $17*68,  for  11  months  28  days,  at  6  per 
cent.?  Ans.  $1'054. 

8.  What  is  the  interest  of  $200,  for  1  day,  at  6  per  cent.  ?  2 

i?  3  days?  4  days?  5  days? 

Ans.  for  5  days,  $0'166. 

9.  What  is  the  interest  of  half  a  mill,  for  567  years,  at  6  per  cent.  ? 

Ans.  $0'017. 

10.  What  is  the  interest  of  $81,  for  2  years  14  days,  at  £  per 
cent.  ?  |  per  cent.  ?   — —  |  per  cent.  ?   2  per  cent.  ?  - 

3  per  cent  ?  4£  per  cent.  ?  5  per  cent.  ?  —  -  6  per  cent.  ? 

7  per  cent.  ?   7£  per  cent.  ?  8  per  cent.  ?   9  per 

cent.?  10  per  cent.?  12  per  cent.?  12£  per  cent.? 

Ans.  to  the  last,  $20' 643. 

11.  What  is  the  interest  of  9  cents,  for  45  years  7  months  11  days, 
at  6  per  cent.  ?  Ans.  $0'246 

12.  A's  note  of  $175  was  given  Dec.  6,  1838,  on  which  was  en 
dorsed  one  year's  interest ;  what  was  there  due  Jan.  1,  1843,  interest 
at  7  per  cent.  ? 

13.  B's  note  of  $56'75  was  given  June  6,  1841,  on  interest  at  6 
per  cent.,  after  90  days ;  what  was  there  due  Feb.  9th,  1842  ? 

Ans.  $58' 197. 

14.  C's  note  of  $365'37  was  given  Dec.  3,  1837  ;  June  7,  1840, 
he  paid  $97' 16;  what  was  there  due  Sept.  11,  1840,  interest  at  5 
per  cent.  ?  Ans.  $318'  184. 

15.  D's  note  of  $203' 17  was  given  Oct.  5,  1838,  on  interest  at  6 
per  cent,  after  3  months  ;  Jan.  5,  1839,  he  paid  $50  ;  what  was  there 
due  May  2,  1841?  Ans.  $174'537. 

16.  E's  note  of  $870'05  was  given  Nov.  17,  1840,  on  interest  at  6 
per  cent,  after  90  days  ;  Feb.  11,  1845,  he  paid  $186'06  ;  what  was 
there  due  Dec.  23,  1847?  Ans.  $1041' 58. 

17.  Supposing  a  note  of  $317*92,  dated  July  5,  1837,  on  which 
were  endorsed  the  following  payments,  viz.,  Sept.  13,  1839,  $208'04  ; 
March  10,  1840,  $76  ;  avhat  was  there  due  Jan.  1,  1841,  interest  at  7 
percent.?  Ans.  $93'032. 

18.  What  will  be  the  annual  insurance,  at  |  per  cent,    on  a  house 
valued  at  $1600?  +  ns.  $10. 

19.  What  will  be  the  insurance  of  a  ship  and  carg:  ,  valued  at 

$5643,  at  \\   per  cent.?    at  |  per  cent.?  at  T7^-  pe.i 

tent.  ?  at  \%  per  cent.  ?  at  J  per  cent.  ? 

Ans.  at  |  per  cent.   $42' 322. 


•  * 

232  PERCENTAGE.  f  179, 

20.  A  man  having-  compromised  with  his  creditors  at  62£  cents  on 
a  dollar,  what  must  he  pay  on  a  debt  of  $137'46?     Arts.  $85'912. 

21.  What  is  the  value  of  $800  stock  in  the  Utica  and  Schenectady 
railroad,  at  1124  per  cent.  ?  Ans.  $900. 

22.  What  is  the  value  of  $5GO'75  of  stock,  at  93  per  cent.  ? 

Ans.  $521'497. 

23.  What  principal,  at  7  percent.,  will,  in  9  months   18  days, 
amount  to  $422'40  ?  Ans.  $400. 

24.  What  is  the  present  worth  of  $426,  payable  in  4  years  and  12 
days,  discounting  at  the  rate  of  5  per  cent.  ? 

In  large  sums,  to  bring  out  the  cents  correctly,  it  will  sometimes  be 
necessary  to  extend  the  decimal  in  the  divisor  to  five  places. 

Ans.  $354'507-{-. 

25.  A  merchant  purchased  goods  for  $250  ready  money,  and  sold 
them  again   for  $300,  payable  in  9  months;  what  did  he  gain,  dis 
counting  at  G  per  cent.  ?  Ans.  $37*081. 

26.  Sold  goods  for  $3120,  to  be  paid,  one  half  in  3  months,  and 
the  other  half  in  0  months  ;  what  must  be  discounted  for  present  pay 
ment,  at  6  per  cent.?  Ana.  $68'491  -f- 

27.  The  interest  on  a  certain  note,  for  1  year  9  months,  at  6  per 
pent.,  was  $49'875  ;  what  was  the  principal?     ':  Ans.  $475. 

28.  What  principal,  at  5  per  cent.,  in  16  months  24  days,  will 
gain  $35?  Ans.  $500. 

29.  Tf  I  pay  $15'50  interest  for  the  use  of  $500,  9  months  and  9 
days,  what  is  the  rate  per  cent.  ? 

30.  If  I  buy  candles  at  $%107  per  lb.,  and  sell  them  at  20  cents, 
what  shall  I  gain  in  laying  out  6100?  Ans.  $19*76. 

31.  Bought  37  gallons  Of  brandy,  at  $1'10  per  gallon,  and  sold  it 
for  $40  ;   what  was  gained  or  lost  per  cent.  ? 

32.  Bought  cloth  at  84 '48  per  yard  ;  how  must  I  sell  it  to  gain 
12A  per  cent.  ?  Ans.  $5'04. 

33.  Bought  50  gallons  of  brandy,  at  92  cents  per  gallon,  but  by 
accident  10  gallons  leaked  out ;  for  what  must  I  sell  the  remainder 
per  gallon,  to  gain  upon  the  whole  cost  at  the  rate  of  10  per  cent.  ? 

Ans.  Sl'265  per  gallon. 

34.  A  merchant  bought  10  tons  of  iron  for  $950  ;  the  freight  and 
duties  were  $145,  and  his  own  charges  $25 ;  how  must  he  sell  it  per 
lb.  to  gain  20  per  cent.  ?  Ans.  6  cents  per  lb. 

35.  A  note  is  jriven  for  $2000,  at  6  per  cent,  annual  interest,  pay 
able  in  6  years ;  the  date  of  the  note  is  Dec.  1,  1841  ;  there  are  en 
dorsements  upon  it  as  follows  :  June  l,iJL842,  $163;  Feb.  1,  1843, 
&12  :  Jan.  1,  1844,  $300  ;  April  1,  1845,  $20;  June  1,  1845,  $20; 
Aug.  1,    1845,   $400;  Jan.    1,    1846,   §100;  Aug.   1,  1847,   $150; 
Oct.  1.  1847,  $75.     What  remained  due,  Dec.  1,   1847,  calculated 
by  the  U.  S.  Court  rule,  by  the  Connecticut  rule,  and  by  the  Ver 
mont  rule,  and  how  do  the  results  compare  ? 

Ans   U.  S..  S1368'81  ;  Ct:,  $1372<5G ;  Vt.,  1274*78. 


11  180,  181.  EQUATION  OF  PAYMENTS.  233 


EQUATION    OF    PAYMENTS. 

1T18O.  1.  A  country  merchant  owes  in  Boston,  $200 
due  in  2  months,  and  $200  due  in  6  months,  each  without 
interest ;  at  what  time  could  he  pay  both  debts,  that  neither 
party  may  lose  ? 

SOLUTION.  —  He  may  keep  the  first  as  long  after  it  is  due  as  he  pays 
the  last  before  it  is  due.  Ans.  4  months. 

The  method  of  finding  the  time  when  several  debts,  due  at 
different  times  without  interest,  should  be  paid  at  once,  is  called 
Equation  of  Payments. 

The  time  of  payment  thus  found,  is  called  the  mean  time. 

2.  A  man  owes  $106,  due  in  one  year,  and  $106,  due  in 
3  years,  without  interest;  in  what  time  shall  he  pay  both  at 
once  ? 

SOLUTION.  —  At  the  end  of  2  years.  But  this,  which  is  the  common 
method,  is  a  gain,  in  this  example,  of  $'36  to  the  debtor.  He  keeps  the 
first  debt  a  year  after  it  is  due,  and  thus  gains  (at  6  per  cent.)  the  inter 
est  on  $106  for  a  year,  or  $6<36.  He  pays  the  whole  of  the  second  debt 
a  year  before  due,  when  he  should  pay  only  such  a  sum  as  would  amount 
to  $106  hi  a  year,  or  $100.  Thus  he  loses  $6  on  the  second  debt,  while 
he  has  gained  S6'36  on  the  first.  The  error,  which  results  from  consid 
ering  the  interest  and  discount  on  the  same  sum  for  the  same  time  equal, 
is  not  usually  regarded  in  business. 

IT  181.      Tofi?id  a  rule  for  the  equation  of  payments. 

1.  Borrowed  of  a  friend  S6'00,  for  4  months  ;  afterwards  1 
lent  him  $1,  to  keep  long  enough  to  balance  the  use  of  the 
money  borrowed  ;  how  long  must  this  be  ? 

SOLUTION.  —  He  should  keep  $1  six  times  as  long,  or, 

Ans.  24  months. 

2.  In  how  long  time  will  $8  be  worth  as  much  as  $40  for 
1  month  ? 

SOLUTION.  —  Every  $8  in  me  $40  will  be  worth  as  much  in  1  month, 
as  the  first  sum,  $8,  is  worth  in  that  time.  Then  as  many  limes  as  $8 
are  contained  in  $40.  so  many  months  the  $8  will  require  to  be  worth 
as  much  as  the  $40  for  1  month.  Ans.  5  months. 

3.  I  have  3  notes   against  a  man  :    1  of  $12,  due  in  3 

Questions.  —  If  ISO.     On  what  principle  is  the  time  of  paying  at  once 
the  two  debts,  Ex.  1,  determined?    What  is  understood  by  the  mean  time 
What  is  equation  of  payments?     What  srror  appears,  Ex.  2,  and  why? 

20* 


234  EQUATION  OF  PAYMENTS.  IT  181 

months;  1  of  $9,  due  in  5  months;  1  of  $6,  due  in  10 
months,  all  without  interest ;  when  should  he  pay  the  whole 
at  once? 

SOLUTION.  —  $12  for    3  months,  is  the  same  as  $36  f;r  1  month. 
$9   "     5  "  "  $45          « 

$6   «  10  "  "  $60          " 

He  has  $2'.  long  enough  to  balance  $141  for  1  month.  Every  $27  in 
$141  will  be  worth  as  much  in  1  month  as  the  first  $27.  Then  as  many 
times  as  $27  is  contained  in  $141,  so  long  he  can  keep  the  $27. 

141 -f- 27  =  5  mo.  6 -{-days,  Ans 

Hence,   To  find  the  mean  time  of  several  payments, 

RULE. 

Multiply  each  sum  by  its  time  of  payment,  and  divide  the 
sum  of  the  products  by  the  sum  of  the  payments. 

EXAMPLES  FOR  PRACTICE. 

4.  A  western  merchant  owes  in  New  York  city  $200,  due  in  5 
months;  $325'50,  due  in  3  months,  and  $413'37,  due  in  2  months; 
but  he  finds  that  it  will  be  more  convenient  to  make  payment  at  one 
time  ;  in  what  time  will  this  be  ? 

Ans.  2'984  months  =  2  months  29  -f-  days. 

5.  I  owe  several  debts,  due  in  different  times,  without  interest, 
namely,  $309'50,  in  8  months  ;  $161,  in  5  months  and  18  days,  and 
$63'25,  in  10  months  and  11  days;  what  shall  I  pay  now  to  cancel 
the  whole,  the  rate  being  6  per  cent.  ? 

NOTE  1.  First  find  the  mean  time,  then  the  present  worth,  IT  164.  The 
fraction  of  a  day  will  not  be  regarded  in  business  operations. 

Ans.  $514*3754-. 

6.  A  merchant  has  owing-  him  $300,  to  be  paid  as  follows :  $50 
in  2  months,  $100  in  5  months,   and  the  rest  in  8  months  :  and  it  is 
agreed  to  make  one  payment  of  the  whole  ;  in  what,  time  ought  that 
payment  to  be1?  Ans.  6  months. 

7.  A  owes  B  $136,  to  be  paid  in  10  months ;  $96,  to  be  paid  in  7 
months  ;  and  $260,  to  be  paid  in  4  months  ;  what  is  the  equated  time 
for  the  payment  of  the  whole  1  Ans.  6  months  7  days  -J-. 

8.  A  owes  B  $600,  of  which  $200  is  to  be  paid  at  the  present 
time,  200  in  4  months,  and  200  in  8  months ;  what  is  the  equated 
time  for  the  payment  of  the  whole?  Ans.  4  months. 

9.  A  owes  B  $300,  to  be  paid  as  follows :  J  in  3  months,  £  in  4 
months,  and  the  rest  in  6  months  ;  what  is  the  equated  time  ? 

Ans.  4£  months. 

NOTE  2.  Sometimes  retailers  sell  on  6  months'  credit,  without  interest. 
But  as  it  would  be  difficult  to  settle  each  item  of  a  long  account  just  6  months 
from  the  time  of  purchase,  all  the  charges  for  a  year  are^setlled  at  its  close. 
Presuming  that  the  purchases  each  month  are  uniform,  this  gives  6  months  as 

Questions. —  IT  H81.  Give  the  solution  of  Ex.  2.  Give  the  solution  of 
Jie  3d  example.  Ruh.  Give  the  substance  of  ihe  notes. 


fi  182,  183.  RATIO.  235 

the  mean  time  of  a  settlement.  Should  a  settlement  be  made  at  the  end 
of  8  months,  the  mean  time  would  be  4  months  ;  at  the  end  of  6  months, 
the  mean  time  would  be  3  months. 


RATIO. 

fl  18&.     How  many  times  is  4  contained  in  8  ? 

Ans.  2  times. 

To  find  how  many  times  one  number  is  contained  in  an 
other,  is  to  find  the  ratio  between  the  numbers,  which  we  do 
by  dividing  one  of  the  numbers  by  the  other.  But  without 
performing  the  division,  we  may  express  it, 

First,  by  the  sign  of  division  :  thus,  8-7-4. 

Second,  by  a  line  without  dots,  writing  the  divi 
dend  in  place  of  the  upper,  and  the  divisor  in  place 
of  the  lower  dot :  thus,  f . 

Third,  by  dots  without  a  line  :  thus,  8:4. 

The  last  is  the  usual  method  of  expressing  ratio,  when  the 
quotient  in  division  receives  this  name.  Hence, 

Ratio  is  the  quotient  expressing  how  many  times  one  num 
ber  is  contained  in  another,  or  how  many  times  one  quantity 
is  contained  in  another  of  the  same  kind. 

NOTE.  —  Ratio  can  only  exist  between  quantities  of  the  same  kind,  since 
the  dividend  and  divisor  must  be  of  the  same  kind.  It  would  be  absurd  to 
inquire  how  many  times  3  bushels  of  rye  are  contained  in  121bs.  of  butter. 
(See  T!  31.)  But  a  ratio  can  exist  between  numbers  of  different  denomina 
tions  when  they  can  be  reduced  to  the  same  denomination ;  thus,  we  can 
determine  how  many  times  8  quarts  are  contained  in  6  gallons,  when  we 
reduce  the  quarts  to  gallons,  or  the  gallons  to  quarts. 

A  ratio  requires  two  numbers,  each  of  which  is  called  a  term  of  the  ratio, 
and  together  they  are  called  a  couplet.  The  first  term,  which  is  the  dividend, 
is  called  the  antecedent ;  the  second,  or  divisor,  is  called  the  consequent. 

Hence  it  follows  that  multiplying  the  antecedent  or  dividing  the  conse 
quent  multiplies  the  ratio,  (11 66,)  dividing  the  antecedent  or  multiplying 
the  consequent,  divides  the  ratio,  (H  57,)  and  multiplying  or  dividing  both 
antecedent  and  consequent  by  the  same  number  does  not  alter  the  ratio. 
(IT  58.) 


Inverted  and  Direct  Ratios. 

H  183.     In   the  ratio  8:4,  the  first  term  is   divided  by 

Questions.  — 11 182.  What  is  it  to  find  the  ratio  of  numbers,  and  how 
done  ?  Describe  the  three  ways  of  expressing  the  division.  What  is  said 
of  the  last  way  ?  Define  ratio.  How  does  the  note  limit  ratio  ?  Why  ? 
Illustrate.  How  many  numbers  a.re.  required,  and  how  many  different 
names  do  they  receive  ?  Apply  IT  56 ;  — .  U 57 ;  —  U  58. 


236  PROPORTION.  IT  184, 185. 

the  second,  and  the  ratio  is  said  to  be  direct.  But  sometimes 
the  second  is  divided  by  the  first,  and  the  rat.o  is  then  said  to 
be  inverse,  since  it  is  equivalent  to  inverting  the  terms,  and 
writing  the  expression  4  :  S.  The  latter  is"  ilso  called  a  re 
ciprocal  ratio,  since  J,  the  quotient  of  4  :  8,  is  the  reciprocal 
of  2,  the  quotient  of  8  :  4,  (11  55.)  Hence, 

Direct  ratio  is  the  quotient  of  the  antecedent  divided  by  the 
consequent;  and 

Inverse  or  reciprocal  ratio  is  the  quotient  of  the  consequent 
divided  by  the  antecedent. 


Compound  Ratio. 

IT  184.  We  have  seen,  IT  79,  that  a  compound  fraction 
consists  of  several  simple  fractions,  to  be  multiplied  together. 
Thus,  the  numerators  of  the  compound  fraction  |  of  J^-,  are 
to  be  multiplied  together  for  a  new  numerator,  and  the  de 
nominators  for  a  new  denominator.  But  since  the  terms  of  a 
ratio  may  be  written  fractionally,  we  may  call  the  expression 
a  compound  ratio. 

Hence,  A  compound  ratio  consists  of  several  simple  ratios  to 
be  multiplied  together,  which  is  done  by  multiplying  together 
the  antecedents,  and  also  the  consequents. 


PROPORTION. 

IT  185.  1.  If  12  yards  of  cloth  cost  $18,  what  will  4 
yards  cost? 

SOLUTION.  —  As  4  yards  are  J  of  12  yards,  they  will  cost  $  of  $18,  or 
$6.  The  12  yards  contain  4  yards  as  many  times  as  $18  contain  $6; 
that  is,  12  -t-  4  =  18  -:-  6  ;  or  fractionally,  -^  =  -^-. 

We  have  here  two  ratios,  which  are  equal.     But  as  the  sign  of  divi 
sion  is  written  to  express  ratio  without  a  line,  4  dots  may  be  written  to 
express  the  equality  of  the  ratios.     The  four  dots  are  used  instead  of 
the  lines  usually  employed.    The  expression  then  becomes 
12  :  4":  :  18  :  6,  equivalent  to  *£-  =  -^-> 

Such  an  expression  is  called  a  proportion,  and  is  rea.d,  12  divided  by 
4  equals  18  divided  by  6  ;  or  more  commonly,  12  is  to  4  as  18  is  to  6. 
Hence, 

Questions.  —  IF  183.  How  does  inverse,  differ  from  direct  ratio  ?  De 
fine  each.  What  other  name  has  inverse  ratio '?  Why  ? 

IT  184.  Whence  arises  compound  ratio?  Defiae  it.  Give  an  example 
Write  it  in  the  common  form.  Reduce  it  to  a  simpfe  ratio. 


H 186.  PROPORTION.  237 

Proportion  is  the  combination  of  two  equal  ratios.  The 
first  and  last  terms  are  called  the  extremes,  the  second  and 
third,  the  means.  The  two  antecedents  are  called  corre 
sponding  terms,  as  are  also  the  two  consequents,  since  these 
terms  have  always  a  certain  reference  to  each  other.  The 
third  term,  $18,  of  this  proportion,  expresses  the  value  of  12 
yards,  the  first  term ;  and  the  fourth  term,  $6,  expresses  the 
value  of  4  yards,  the  second  term. 

NOTE. — A  proportion  requires  four  terms,  two  antecedents  and  two  conse 
quents.  Three  numbers  may  form  a  proportion  if  one  is  used  in  both  ratios  ; 
thus,  with  the  numbers  12,  6,  and  3,  we  have  the  proportion,  12 :  6 : :  6 : 3. 


Rule   of  Three. 

IT  186.  When,  as  in  the  above  example,  the  first  three 
terms  are  given  to  find  a  fourth,  we  may  find  it  by  taking 
such  a  part  of  the  third  term  as  the  second  is  of  the  first ;  or 
by  a  method  called  the  Rule  of  Three,  on  the  following  prin 
ciples.  Take  the  proportion, 

12  :  4  : :  IS  :  6,  fractionally  expressed,  -^-=  -^. 

As  the  fractions  are  equal,  if  we  reduce  them  to  the  common 
denominator,  24,  by  the  rule,  IF  70,  the  numerators  will  be 
equal,  and  the  fractions  will  become,  £f  =  Jf- 

The  first  numerator,  72,  it  may  be  seen,  is  the  product  of 
the  extremes,  and  the  second  numerator,  72,  is  the  product 
of  the  means,  of  the  proportion.  Hence, 

The  product  of  the  extremes  of  a  proportion  is  equal  to  the 
product  of  the  means. 

Take  the  first  thrct;  terms  to  find  the  fourth. 

yds.  yds.     doll.     doll.  SOLUTION.  —  We  multiply  together 

12  I  4  1 1  18  I   •  •  •  •  4  and  18,  the  means,  which  gives  the 

4  product  of  the  extremes,  of  which  one 

extreme,  12,  is  given,  and  we  divide 

12 )  72  the  product  by  the  given  factor  to  get 

the  other  extreme,  or  fourth  term. 
6  dollars,  Ans. 

Questions.  —  IT  185.  How  is  the  price  of  the  4  yards,  Ex.  1,  found? 
What  two  ratios  are  formed  ?  By  what  sign  is  their  equality  expressed,  and 
instead  of  what  is  it  used?  Give  the  two  ways  of  reading  a  proportion.  De 
fine  proportion.  What  are  its  terms?  What  are  extremes?  What  are 
means  ?  What  are  corresponding  terms,  and  why  ?  How  many  terms,  and 
how  many  numbers,  are  requir*  1  in  proportion  1  Illustrate. 


238  PROPORTION.  H 187, 188. 

NOTE.  — This  operation  is  strictly  analytic,  for  had  the  price  of  1  yard  been 
18  dollars,  we  should  have  multiplied  it  by  4  to  get  the  price  of  4  yards.  But 
as  it  is  12  yards,  which  cost  18  dollars,  the  product  of  18  by  4  is  12  times  too 
large,  and  must  be  divided  by  12. 

5T  1 871.     To  write  down  the  three  given  numbers. 

2.  At  $90  for  15  barrels  of  flour,  how  many  barrels  can  be 
bought  for  $30  ? 

doll.  doll.      bar.      bar.  SOLUTION. — We  have  the  terms  of  one 

90  *  30  ::  15  :   ••••  ratio,  $90  and  $30,  being  of  the  same 

30  kind.    For  the  same  reason,  15  barrels 

and  the  required  number  of  barrels  will 

9 1 0 )  45 1  0  form  another  ratio.    We  place  the  ante- 

cedent  of  the  second  ratio,  15  barrels,  for 

5  bar.,  Ans.       the  third  term,  and  its  correspondent,  $90 
for  the  first,  and  $30  for  the  second,  that 

its  correspondent  may  be  the  fourth  term.   We  then  multiply  and  divide 
as  above,  and  get  the  Ans.,  5  barrels. 

IF  188.      To  invert  both  ratios. 

In  the  proportion,  90  :  30  : :  15  :  5,  the  quotient  of  90  -4- 
30  is  3,  and  that  of  15  -f-  5  is  3.  Inverting  the  terms  of  each 
couplet,  we  still  have  the  proportion,  30  :  90  : :  5  :  15,  for 
each  ratio  is  now  J,  the  reciprocal  of  the  former  ratio,  (II 183,) 
and  consequently  the  two  ratios  are  equal.  This  inversion 
often  virtually  occurs  in  operations ;  thus, 

3.  At  $30  for  5  barrels  of  flour,  how  many  barrels  can  be 
bought  for  $90  ? 

doll.    doll.       bar.    bar. 

90  :  30  : :  ••  ••  5          SOLUTION.  —  Writing  the  first  ratio,  90 :  30, 

By  inversion.  as  above,  5  barrels  must  be  the  fourth  term, 

30  *  90  •  •  5  •    ....     as  il  corresponds  to  $30,  the  second  term. 

_  '  *  But  it  is  convenient  always  to  regard  the 

fourth  as  the  unkrh  'vn  term,  and  it  will  be 

Q  I  n  1>  A^  I  n  come  so  bv  inverting  both  ratios,  when  the 

operation  is  the  same  as  before. 

15  bar.,  Ans.  Ans-  15  barrels- 


Questions.  —  T  186.  What  is  the  object  of  the  rule  of  three  ?  Com- 
pare  the  two  methods  of  expressing  a  proportion,  and  show  what  results  from 
reducing  the  fractions  to  a  common  denominator.  How  is  the  first  numerator 
obtained  1  —  the  second  ?  What  important  conclusion  is  derived  ?  How  ap 
plied  to  finding  the  fourth  term  ?  How  explained  analytically  ? 

V  187.  What  must  be  made  the  third  term  ?  Why  ?  What  must  be 
made  the  first,  and  why  ?  What  the  second  ?  Why  ? 

jj,  88.  Why  can  both  ratios  be  inverted?  What  is  the  object  in  so 
doing  ? 


H  189,  190.  PROPORT  ON.  239 

IT  1  8O«     1  >  invert  one  ratio. 

4.    If  3  men  will  build  a  wall  in  10  days,  in  how  long  time 
will  6  men  build  it  ? 

•men.  men.    days.    days.  SOLUTION.  —  "Writing    the    corre- 

3   '    6  '  *  10  !   ••••  spending  terms  as  already  described, 


the  tot  ratio,  we  have, 

6    I    3  :  :  10  time  required  by  3  men,  and  the  un- 

3  known  term  the  time  required  by  6 

men.     But  since  twice  the  number 

/*  \  on  of  men  will  build  the  wall  in  half 

the  time,  3  men  are  such  a  part  ot 

(are  contained  in)  6  men  as  the  re 

5  days,  Ans.  quired  number  of  days  are  a  part  of 

10  days.     The  first,  then,  is  an  in 

verse  ratio,  the  consequent  being  divided  by  the  antecedent.  But  if  we 
invert  its  terms,  the  division  will  be  as  usual,  and  the  operation  as  al- 
leady  described. 

NOTE.  —  In  the  third  example,  more  money  would  buy  more  flour;  in  the 
second,  less  money  would  buy  less  flour.  In  such  examples,  each  antecedent 
is  divided  by  its  consequent,  or  if  one  ratio  is  inverted,  the  other  is  also.  The 
proportion  is  then  called  direct.  But  when,  as  in  the  fourth  example,  more 
requires  less,  (more  days,  less  time,)  or,  as  may  be  the  case,  less  requires 
more,  one  ratio  is  inverse,  while  the  other  is  direct.  The  proportion  is  then 
called  inverse.  Hence, 

Direct  proportion  is  the  combination  of  two  direct,  or  two 
inverse  ratios. 

Inverse  proportion  is  the  combination  of  a  direct  and  an 
inverse  ratio.  But  if  one  of  the  ratios  is  inverted,  it  may  be 
treated  as  a  direct  proportion. 

If  1OO.  Hence,  to  find  the  fourth  term  of  a  proportion 
when  three  terms  are  given,  we  have  the  following 

RUJLE. 

I.  Make  that  one  of  the  three  numbers  the  third  term 
which  is  of  the  same  kind  as  the  answer  sought,  reducing  the 
other  two,  if  necessary,  to  the  same  denomination,  that  the 
terms  of  each  ratio  may  be  of  the  same  kind. 

II.  Write  that  one  of  the  remaining  two  for  the  first  term 
which  corresponds  to  the  third,  and  the  other  for  the  second 
term. 


Questions.  —  1T 189.  How  are  the  numbers,  Ex.  4,  to  be  written  accord 
ing  to  the  rule  for  the  corresponding  terms?  What  happens  from  this  man 
ner  of  writing  them,  and  whv?  What  then  is  done?  How  do  the  second 
and  third  differ  from  the  fourth  example  ?  What  is  direct  proportion  ?  What 
inverse  ? 


240  PROPORTION.  IF  191. 

III.  If  the  question  involves  the  inverse  proportion,  more 
requiring  less,  or  less  requiring  more,  invert  the  first  ratio. 

IV.  Multiply  together  the  second  and  third  terms,  or  two 
means,  to  get  the  product  of  the  extremes,  which  being  divided 
by  the  first  term,  or  known  extreme,  the  quotient  will  be  the 
other  extreme,  or  fourth  term  sought,  of  the  same  kind  as  the 
third  term. 

NOTE.  —  If  the  third  term  is  a  compound,  or  a  mixed  number,  it  must  be 
made  of  but  one  denomination. 

EXAMPLES    FOR    PRACTICE. 

^T  1O1.  1.  If  6  horses  consume  21  bushels  of  oats  in  3 
weeks,  how  many  bushels  will  serve  20  horses  the  same 
time  ? 

horses,  horses,    oats.      oats. 

2                         7  NOTE  1  .  —  Since  in  the  operation  there  are 

0.     OH  ••  &1  •    .  always  two  numbers  to  be  multiplied  to- 

•     *-'-'  •  •  f&A-  •    •  gether,  and  their  product  to  be  divided  by  a 

7  third  number,  the  process  may  frequently  be 

_  shortened   by  cancelation,  as  shown  ;  "  the 

n\-\Af\  factor,  3,  being  canceled  in  21  and  6,  and 

<Z  )  l^tU  jjje  remaining  factors,  7  and  2,  being  used  as 

multiplier  and  divisor. 
70  bush.,  Ans. 

2.  The  above  question  reversed.     If  20  horses  consume  70  bushels 
of  oats  in  3  weeks,  how  many  bushels  will  serve  6  horses  the  same 
time?  Ans.  21  bushels. 

3.  If  365  men  consume  75  barrels  of  provisions  in  9  months,  how 
much  will  500  men  consume  in  the  same  time  ? 

73  15 


Ans.  1024f  barrels. 

4.  If  500  men  consume  102^f  barrels  of  provisions  in  9  months, 
how  much  will  365  men  consume  in  the  same  time  ? 

100         73 
$00  :  $0$  :  :  102^f  :    ••  ••      And  reducing  the  third  tern,  to  an 

7^00 
improper  fraction,  we  have.  100  :  Jf#  I  :  —  -  :   ••  ••     Now,  since 

Tip 

the  denominator  is  a  divisor,  (^[  64,)  we  cancel  73,  and  have,  100  i 
1  :  :  7500  :  ••  ••  Ans.  75  barrels. 

5.  If  the  moon  move   13°  10'  35"  in  1  day,  in  what  time  does  it 
perform  one  revolution?  Ans.  27  days  7  h.  43  m.  6  s.  -}-• 

Questions.  —  IT  190.  What  number  is  made  the  third  term,  and  why  ? 
How  are  the  other  two  numbers  arranged  ?  When  is  the  first  ratio  inverted  1 
For  what  is  the  multiplication?  —  the  division?  When  is  a  reduction 
needed  ?  Repeat  the  whole  rule. 


IT  191.  PROPORTION.  241 

6.  If  a  person,  whose  rent  is  $145,  pay  $12'63  parish  taxes,  how 
much  should  a  person  pay  whose  rent  is  $378  ?          Ans.  $32'925. 

7.  If  I  buy  7  Ibs.  of  sugar  for  75  cents,  how  many  pounds  can  I 
buy  for  $6  ?  Ans  56  Ibs. 

^  NOTE  2.  —  Every  example  in  proportion  may  be  performed  on  general  prin 
ciples,  IT  134.  Thus,  in  the  last  example,  we  may  divide  $'75,  the  price  of  7 
Ibs.,  by  7,  ami  we  shall  have  the  price  of  1  lb.,  and  then  divide  $&-,  the  price 
of  the  required  number  of  Ibs.,  by  the  price  of  1  lb.  now  found,  and  it  will  give 
us  the  number  of  Ibs. 

8.  If  I  give  $6  for  the  use  of  $100  for  12  months,  what  must  I 
give  for  the  use  of  $357'82  the  same  time?  Ans.  $21'469-{- 

On  general  principles.  Divide  $6  by  100  to  get  the  gain  on  $1, 
which  multiply  by  the  number  of  dollars,  to  get  the  gain  on  the  num 
ber  of  dollars. 

NOTE  3.  — Let  the  pupil  be  required  to  perform  all  the  subsequent  examples 
both  by  proportion  and  on  general  principles,  or  analysis. 

9.  If  a  staff  5  ft.  8  in.  in  length,  cast  a  shadow  of  6  feet,  how  high 
is  that  steeple  whose  shadow  measures  153  feet?     Ans.  144£  feet. 

10.  If  a  family  of  10  persons  use  3  bushels  of  malt  in  a  month, 
how  many  bushels  will  serve  them  when  there  are  30  in  the  family  ? 

Ans.  9  bushels. 

By  Analysis.     If  10  persons  use  3  bushels,  1  person  will  use  -^ 
of  3  bushels,  or  T^j-  of  a  bushel.     And  if  1  person  use  -j^j-,  30  per 
sons  will  use  30  times  .3%.=  f-$=9  bushels. 
NOTE.  4  — The  seven  following  examples  iavolve  the  inverse  proportion. 

11.  There  was  a  certain  building  raised  in  8  months  by  120  work 
men  ;  but  the  same  being  demolished,  it  is  required  to  be  built  in  2 
months ;  I  demand  how  many  men  must  be  employed  about  it. 

Ans.  480  men. 

12.  There  is  a  cistern  having  a  pipe  which  will  empty  it  in  10 
hours ;  how  many  pipes  of  the  same  capacity  will  empty  it  in  24  min 
utes?  Ans.  25  pipes. 

<  13.  A  garrison  of  1200  men  has  provisions  for  9  months,  at  the 
rate  of  14  oz.  per  day  ;  how  long  will  the  provisions  last,  at  the  same 
allowance,  if  the  garrison  be  reinforced  by  400  men? 

Ans.  61  months. 

14.  If  a  piece  of  land   40  rods  in  length  and  4  in  breadth,  make 
an  acre,  how  wide  must  it  be  when  it  is  but  25  rods  long? 

Ans.  6§  rods. 

15.  If  a  man  perform  a  journey  in  15  days,  when  the  days  are  12 
hours  long,  in  how  many  will  he  do  it  when  the  days  are  but  10  hours 
long?  Ans.  18  days. 

16.  If  a  field  will  feed  6  cows  91  days,  how  long  will  it  feed  21 
cows?  Ans.  26  days. 

Questions.  —  IT  191.  How  is  the  operation  in  proportion  shortened? 
How  may  every  example  in  proportion  be  performed  without  stating  it? 
What  is  the  method  described,  IT  134,  for  examples  requiring  several  opera 
tlons  ? 

21 


242  PROPORTION.  f  192. 

17.  Lent  a  friend  292  dollars  for  6  months;  some  time  after,  he 
.ent  me  806  dollars  ;  how  long  may  I  keep  it  to  balance  the  favor  1 

Ans.  2  months  5-}-  days. 

18.  If  7  Ihs.  of  sugar  cost  |  of  a  dollar,  what  cost  12  Ibs.  ? 

Ans.  $lf. 

19.  If  6^  yds.  of  cloth  cost  $3,  what  cost  9|  yds? 

Ans.  $4<269-[~. 

20.  If  2  oz.  of  silver  cost  $2'24,  what  costs  |  oz.  ? 

Ans.  $0*84 

21.  Iff  oz.  cost  $-y.,  what  costs  1  oz.  ?  Ans.  $1'283. 

22.  Iff  yd.  cost  $£,  what  will  40  J  yds.  cost?  Ans.  $59'062-|-. 

23.  A  merchant,  owning  ^  of  a  vessel,  sold  §  of  his  share  foi 
$957  ;  what  was  the  vessel  worth?  Ans.  $1794'375. 

••  24.  If  12  acres  3  roods  produce  78  quarters  3  pks  of  wheat,  how 
much  will  35  acres  1  rood  20  poles  produce? 

f  Ans.  216  qrs.  5  bush.  1  pk.  4  qts. 

-  25.  A  cistern  has  4  pipes  which  will  fill  it,  respectively,  in  10,  20, 
40,  and  80  minutes ;  in  what  time  will  all  four,  running  together,  fill 
it  ?  Ans.  5$  minutes. 

26.  At  $33  for  6  barrels  of  flour,  what  must  be  paid  for  178  bar 
rels?  Ans.  $979. 

27.  If  2'5  Ibs.  of  tobacco  cost  75  cents,  how  much  will  185  Ibs. 
cost?  Ans.  $55'50. 

28.  What  is  the  value  of  '15  of  a  hogshead  of  lime,  at  $2'39  per 
hhd.?  Ans.  $0'3585. 

29.  If  '15  of  a  hhd.  of  lime  cost  $0'3585,  what  is  it  per  hhd.  ? 

Ans.  $2 '39. 

30.  A  bankrupt   owes    $972,    and   his   property,    amounting   to 
$607'50,  is  distributed  among  his  creditors  ;  what  does  one  receive 
whose  demand  is  $11J?  Ans.  $7*083 +• 

31.  When  wheat  is  worth  $'93  per  bushel,  a  3  cent  loaf  weighs 
12  oz.  ;  what  must  it  weigh  when  wheat  is  $1'24  per  bushel? 

Ans.  9  oz. 

32.  A  company  of  16  men  are  on  an  allowance  of  6  oz.  of  bread  a 
day ;  what  will  be  their  daily  allowance  for  28  days,  if  they  receive 
an  addition  of  224  Ibs.  ?  Ans.  14  oz. 

f?% 


Compound  Proportion. 

IT  192.  1.  If  a  man  travel  240  miles  in  8  days  of  12  hours 
long,  how  far  will  he  travel  in  6  days  of  10  hours  long,  trav 
eling  at  the  same  rate  ? 

SOLUTION.  —  The  relation  of  the  number  of  miles,  240,  which  must  be 
iflade  the  third  term,  to  the  required  distance,  depends  on  two  circum 
stances,  the  number,  and  the  length  of  the  days. 


1T 192.  PROPORTION.  243 

First,  considering  the  number  of  days,  without  regard  to  Ineir  length, 
vre  shall  have, 

days.  days,      miles.        miles. 

8:6::  240  :  .... 

"We  find  that  he  will  travel  180  miles  in  6  days  of  the  same  length  as 
the  8  days. 

Second,  considering  the  length  of  the  days,  without  regard  to  their 
number,  we  shall  have, 

hours,   hours,      miles.      miles. 

12  :  10  ::  180  :  .... 

He  will  travel  150  miles  in  6  days,  10  hours  long,  vhen  he  travels 
180  in  the  same  number  of  days,  12  hours  long. 

By  the  dependence  of  the  distance  on  the  ratio  8 : 6,  we  get  f  of  240, 
and  on  the  ratio  12 :  10,  we  get  |£  of  this  |.  But  $$  of  I  is  a  com 
pound  fraction,  and  consequently,  8 :  6  and  12 :  10,  on  which  the  dis 
tance  depends,  constitute  a  compound  ratio,  which  may  be  united  with 
the  given  distance,  as  follows : 

The  compound  may  be 


!Q 
12 
Reduced.          96 


>  *  *  240  '   •  reduced  to  a  simple  ra- 

10  J  "  *  tio,  after  which  the  op- 

60      *  *  240  '  •  •  » •        eration  is  the  same  as 


in  a  simple  proportion 
Such  a  proportion  is  called  a  compound  proportion.     Hence, 

A  compound  proportion  is  the  combination  of  a  compound 
and  a  simple  ratio,  and  exists  when  the  relation  of  the  given 
quantity  to  the  required  quantity  of  the  same  kind  depends 
on  several  circumstances. 

NOTE  1.  — That  the  compound  does  not  differ  from  the  simple  proportion, 
may  be  seen  from  the  fact  that,  in  the  above  reduced  proportion,  96  is  the 
number  of  hours  in  which  240  miles  are  traveled,  and  60  the  number  ot  hours 
in  which  the  required  distance  is  traveled. 

2.  If  264  men,  in  5  days  of  12  hours  each,  can  dig  a 
trench  240  yards  long,  3  wide,  and  2  deep,  in  how  many 
days,  of  9  hours  long,  will  24  men  dig  a  trench,  420  yards 
long,  o  wide,  and  3  deep  ? 

SOLUTION.  — The  number  of  days  is  required,  and  its  relation  to  the 
given  quantity,  5,  depends  on  five  circumstances,  the  number  of  men, 
the  length  of  the  days,  the  length,  breadth,  and  depth  of  the  trench. 
Placing  5  for  the  third  term,  we  connect  it  with  a  compound  ratio  com 
posed  of  five  simple  ratios ;  thus, 

Questions.  —  IT  192.  Why  two  operations  to  Ex.  1?  Describe  the 
first ;  —  the  second.  How  does  it  appear  that  the  relation  of  the  distances 
depends  upon  a  compound  ratio  ?  What  is  a  compound  proportion  ?  How  is 
it  seen  that  it  does  not  differ  in  principle  from  the  simple  proportion?  What 
reduction  is  made  on  the  compound  ratio  ?  How  is  the  operation  of  reducing 
shortened  ?  Give  the  rule. 


244  PROPORTION.  IF  193 

11 

Inverse,     &£ 
3 
I?iverse,       0   ;      J$         days-    ^V- 

::  5  :  •••• 

2  Y 

Direct,    &£0 
Direct,         & 
Direct,         2 
3x2  =  6.         11  x?X5  =  385. 

Each  simple  ratio  is  from  one  circumstance,  regarding  the  other  cir 
cumstances  uniform.  The  first  two  ratios,  it  may  be  seen,  are  inverse 
since  the  less  the  number  of  men  employed,  or  the  shorter  the  days, 
the  greater  will  be  the  number  of  days. 

Reducing  this  compound  ratio  to  a  simple  one,  shortening  the  process 
by  cancelation,  we  have  the  simple  proportion,  — 

6  :  3S5 : :  5 

_  By  this  proportion  we  get  the  answer  in  the 

6 )  1925  same  manner  as  by  any  simple  proportion. 

320£  days,  Ans. 

5T  193.  Hence,  questions  tnvolving  a  compound,  propor 
tion,  may  be  performed  by  the  following 

RULE. 

I.  Having  made  that  number  the  third  term  which  is  of 
the  same  kind  as  the  answer  sought,  we  form  a  ratio,  either 
direct  or  inverse,  as  required,  from  the  two  remaining  num 
bers  which  are  of  the  same  kind,  and  place  it  for  one  couplet 
of  the  compound  ratio.     Form  each  two  like  remaining  num 
bers  into  a  ratio,  and  so  on,  till  all  are  used. 

II.  Having  reduced  the  compound  to  a  simple  ratio,  by 
multiplying  together  the   antecedents  and  the  consequents, 
shortening  the  process  by  cancelation,  the  operation  will  be 
the  same  as  in  a  simple  proportion. 

EXAMPLES   FOR    PRACTICE. 

1.  If  6  men  build  a  wall  20  ft.  long,  6  ft.  high,  and  4  ft.  thick,  in 
16  days,  in  what  time  will  24  men  build  one  200  ft.  long,  8  ft.  4iigh, 
and  6  ft.  thick?  Ans.  80  days. 

2.  If  the  freight  of  9  hhds.  of  sugar,  each  weighing  1200  Ibs.,  20 
miles,  cost  $  16,  what  must  be  paid  for  the  freight  of  50  tierces,  each 
weighing  250  Ibs.,  100  miles. 

Questions.  —  IT  193.  What  is  the  rule  for  compound  proportion  ?  How 
many,  and  what  circumstances  in  Ex.  2  ?  How  else  may  the  examples  be 
performed  ? 


1T  194  PROPORTION.  245 

NOTE  ..  —  The  price  of  freight  depends  on  three  circumstances,  the  numbet 
of  casks,  the  size  of  the  casks,  and  the  distance. 

Ans.  $92'59-{-. 

3.  If  56  Ibs.  of  bread  be  sufficient  for  7  men  14  days,  how  much 
bread  will  serve  21  men  3  days?  Ans.  36  Ibs. 

The  same  by  analysis.  If  7  men  consume  56  Ibs.,  1  man  will  con 
sume  \  of  56  =  8  Ibs.  in  14  days,  and  -^  of  8  Ibs.,  =  T?4  of  a  lb., 
in  1  day.  21  men  will  consume  21  times  what  1  man  will  consume, 
that  is,  21  times  T8¥  =  J^-=  12  Ibs.  in  1  day,  and  3  times  12  Ibs 
=  36  Ibs.  in  3  days. 

NOTE  2.  —  Having  wrought  the  following  examples  by  proportion,  let  the 
pupil  be  required  to  do  the  same  by  analysis. 

4.  If  4  reapers  receive  $11'04  for  3  days'  work,  how  many  men 
may  be  hired  16  days  for  $103'04?  Ans.  7  men. 

5.  If  7  oz.  8  drs.  of  bread  be  bought  for  $'06  when  wheat  is  $'76 
per  bushel,  what  weight  of  it  may  be  bought  for  $'18  when  wheat  is 
$'90  per  bushel?  Ans.  1  lb.  3  oz. 

6.  If  $100  gain  $6  in  1  year,  what  will  $400  gain  in  9  months  ? 

NOTE  3.  —  This  and  the  three  following  examples  reciprocally  prove  each 
other. 

7.  If  $100  gain  $6  in  1  year,  in  what  time  will  $400  gain  $18  ? 

8.  If  $400  gain  $18  in  9  months,  what  is  the  rate  per  cent,  per 
annum  ? 

9.  What  principal,  at  6  per  cent,  per  annum,  will  gain  $18  in  9 
months  ? 

10.  A  usurer  put  out  $75  at  interest,  and  at  the  end  of  8  months, 
received,  for  principal  and  interest,  $79  ;  I  demand  at  what  rate  per 
cent,  he  received  interest.  Ans.  8  per  cent. 


f  194.    Review  of  Proportion. 

Questions.  —  What  is  ratio?  Between  what  quantities  does  it 
exist?  What  is  inverse  ratio?  — compound  ratio?  What  is  propor 
tion  ?  —  rale  of  three  ?  How  is  it  shown  that  the  product  of  the  ex 
tremes  equals  the  product  of  the  means  ?  What  inversions  of  the  ratios 
take  place  ?  How  is  it  shown  that  the  operation  of  the  rule  of  three 
depends  on  analysis?  What  is  inverse  proportion?  — direct  propor 
tion  ?  —  compound  proportion  ? 

EXERCISES. 

1.  If  1  buy  76  yds.  of  cloth  for  $11347,  what  does  it  cost  per  ell 
English?  Ans.  $l'861-[-. 

2.  Bought  4  pieces  of  Holland,  each  containing  24  ells  English, 
for  $96  ;  how  much  was  that  per  yard?  Ans.$0l80. 

3.  A  garrison  had  provision  for  8  months,  at  the  rate  of  15  ounces 

21* 


246  ALLIGATION.  IT  195 

to  each  person  per  day ;  how  much  must  be  allowed  per  day,  in  order 
that  the  provision  may  last  9|  months?  Ans.  12-ff  oz. 

4.  How  much  land,  at  $2*50  per  acre,  must  be  given  in  exchange 
for  360  acres,  at  $3'75  per  acre?  Ans.  540  acres. 

5.  Borrowed  185  quarters  of  corn  when  the  price  was  19s. ;  ho\t 
much  must  I  pay  when  the  price  is  17s.  4d.  ?       Ans.  202^  qrs. 

6.  A  person,  owning  •§•  of  a  coal  mine,  sells  |  of  his  share  for 
j£l71 ;  what  is  the  whole  mine  worth?  Ans.  j£380. 

7.  If  f  of  a  gallon  cost  f  of  a  dollar,  what  costs  ^  of  a  tun  ? 

Ans.  $140. 

8.  At  £\\  per  cwt.,  what  cost  31  Ibs.  ?  Ans.  lOfd. 

9.  If  4£  ewt.  can  be  carried  36  miles  for  35  shillings,  how  many 
pounds  can  be  carried  20  miles  for  the  same  money  1 

Ans.  907i  Ibs. 

10.  If  the  sun  appears  to  move  from  east  to  west  360  degrees  in  24 

hours,  how  much  is  that  in  each  hour?  in  each  minute?  

in  each  second?  Ans.  to  the  last,  15"  of  a  deg. 

11.  If  a  family  of  9  persons  spend  $450  in  5  months,  how  much 
would  be  sufficient  to  maintain  them  8  months,  if  5  persons  more  were 
added  to  the  family  ?  Ans.  $1120. 


ALLIGATION— Medial. 

IT  195.  1.  A  farmer  mixed  8  bushels  of  corn,  worth  60 
cents  per  bushel,  4  bushels  of  rye,  worth  80  cents  per  bushel, 
and  4  bushels  of  oats,  worth  40  cents  per  bushel ;  what  was 
a  bushel  of  the  mixture  worth  ? 

When  several  simples  of  different  values  are  to  be  mixed, 
the  process  of  finding  the  average  price,  is  called  Alligation 
Medial.  The  average  price  is  called  the  mean  price. 

OPERATION.  SOLUTION.  —  Multiply 

8  bushels,  at  $<60  ;  60  X  8  =  $4'80.    ™S.  ^  lhe  P™*'*3 

AT        1    7          j.  df  on       on  A  am  ctf\        bUSnel   Ol    COm,  DV    O,  the 

4  bushels,  at  S'80 ;  SO  X  4  =  S3'20.         Ulct  is  the     fce  'of  8 
_4  bushels,  at  $<40  ;   40x4=  $1'60.     bushels.     In  like  manner 

in    i        7     7  7  rthn  nr\        we    nnd    tne    pfJCC   of   the 

16  bushels  are  ivorth  $9'60.    rye>  and  the  oats  in  the 

1  bushel  is  ivorth  $'60.     mixture,  and  adding  to 

gether  the  prices  of  the 

corn,  the  rye,  and  the  oats,  we  have  $9'60,  the  price  of  8  -f-  4  -[-  4  =  16 
bushels,  which  are  contained  in  the  whole  mixture,  and  dividing  the 
price  of  16  bushels  by  16,  we  get  the  price  of  1  bushel.  Ans.  60  cents. 

Hence,  the 

Questions.  —  Tf  195.  What  is  alligation  medial ?  --mean  rate ?  What 
is  the  rule  ?  To  what  does  it  apply  ? 


T  196.  ALLIGATION.  247 

RULE. 

Find  the  prices  of  the  several  simples,  and  add  them  to 
gether  for  the  price  of  the  whole  compound,  which  divide  by 
the  number  of  pounds,  bushels,  &c.,  to  get  the  price  of  1 
pound  or  bushel. 

]\JOTE.  —  The  principles  of  the  rule  are  applicable  to  many  examples  not 
embraced  by  the  above  definition  of  Alligation  Medial. 

EXAMPLES   FOR    PRACTICE. 

2.  A  grocer  mixed  5  Ibs.  of  sugar  worth  10  cents  per  lb.,  8  Ibs. 
worth  12  cents,  20  Ibs.  worth  14  cents  ;  what  is  a  pound  of  the  mix 
ture  worth?  Ans.  $'12-}-°-. 

3.  A  goldsmith  melted  together  3  ounces  of  gold  20  carats  fine, 
and  5  ounces  22  carats  fine  ;  what  is  the  fineness  of  the  mixture  ? 

Ans.  21i  carats. 

•  4.  A  grocer  puts  6  gallons  of  water  into  a  cask  containing  40  gal 
lons  of  rum,  worth  42  cents  per  gallon  ;  what  is  a  gallon  of  the  mix 
ture  worth  1  Ans.  36£f  cents. 

5.  On  a  certain  day  the  mercury  was  observed  to  stand  in  the  ther 
mometer  as  follows :  5  hours  of  the  day  it  stood  at  64  degrees ;  4 
hours,  at  70  degrees ;  2  hours,  at  75  degrees ;  and  3  hours  at  73  de 
grees  ;  what  was  the  mean  temperature  of  that  day  1 

Ans.  C9T3j-  degrees. 

6.  A  farm  contains  16  acres  of  land  worth  $90  per  acre,  22  acres 
worth  $75,  18  acres  worth  $64,  10  acres  worth  $55,  30  acres  worth 
$36,  and  42  acres  worth  $25  per  acre  ;  what  is  the  average  value  of 
the  farm  per  acre?  Ans.  $50' 16  nearly  per  acre. 

7.  A  dairyman  has  20  cows,  3  of  which  are  worth  $35  each,  4  are 
worth  $30,  6  are  worth  $24,  4  are  worth  $20,  2  are  worth  $18  each, 
and  1  is  worth  $13  ;  what  is  the  average  value?         Ans.  $24;90. 


Alligation  Alternate. 

^f  196.  1.  A  farmer  has  1  bushel  of  corn,  worth  50 
cents.  How  many  bushels  of  oats,  worth  40  cents,  must  be 
put  with  it,  to  make  the  mixture  worth  42  cents  ? 

SOLUTION. — The  1  bushel  of  corn  is  worth  8  cents  more  than  the 
prict  of  the  mixture,  and  as  each  bushel  of  oats  is  worth  2  cents  less,  he 
must  take  4  bushels  of  oats  Ans. 

When  the  price  of  several  simples,  (corn  and  oats,)  and  the 
price  of  a  mixture  to  be  formed  from  them  are  given,  the 
method  of  finding  the  quantity  of  each  simple  is  called  Alii" 
gation  Alternate. 


248  ALLIGATION.  f  196. 

2.  How  many  bushels  of  oats  must  the  farmer  take  to  mix 
with  2  bushels  of  corn,  prices  the  same  as  above  ? 

SOLUTION.  —  Evidently,  twice  as  many  as  were  required  for  1  bushel, 
or,  Ans.  8  bushels. 

NOTE  1.  — We  see  that  2,  the  number  of  bushels  of  corn,  equals  tlie  number 
of  cents  that  the  oats  are  worth  less  than  the  mixture,  and  8,  the  number  of 
bushels  of  oats,  equals  the  number  of  cents  that  the  corn  is  worth  more  than 
the  mixture. 

Then,  if  the  three  prices  had  been  given,  we  might  have  taken  2,  the  differ 
ence  between  the  price  of  the  oats  and  of  the  mixture  for  the  number  of  bushels 
of  corn,  and  8,  the  difference  between  the  price  of  the  corn  and  of  the  mixture, 
for  the  number  of  bushels  of  oats,  and  we  should  have  such  a  mixture. 

That  is,  we  take  the  difference  between  the  price  of  each  sim 
ple  and  of  the  mixture  for  the  number  of  bushels  of  the  other 
simple. 

NOTE  2.  —  By  this  process,  the  sum  of  the  excesses,  found  by  multiplying  8 
cents,  the  excess  of  1  bushel  of  corn,  by  2,  the  number  of  bushels,  equals  the 
sum  of  the  deficiencies,  found  by  multiplying  2,  the  deficiency  of  1  bushel  of 
oats,  by  8,  the  number  of  bushels.  Or,  8  X  2  =  2  X  8,  since  the  factors  are  the 
same  in  each. 

3.  A  merchant  has  two  kinds  of  sugar,  worth  6  cents  and 
17  cents  per  pound,  of  which  he  would  make  a  mixture  worth 
10  cents  ;  how  much  of  each  must  he  take  ? 

OPERATION.  SOLUTION. — Take  the 

T-,  .        f     .  A          nri  (     6  n  7  Ibs.     difference   between   10 

Price  of  mixture,  10  cents,  j  1?J  4  lbg      and  17  for  the  number 

of  Ibs.  at  6  cents,  and 

the  difference  between  6  and  10  for  the  number  of  Ibs.  at  17  cents.  The 
sum  of  the  deficiencies,  7  times  4  cents,  equals  the  sum  of  the  excesses, 
4  times  7  cents. 

NOTE  3.  —  This  gives  a  mixture  of  1 1  pounds,  and  if  3  times,  5  times,  one 
half,  or  any  other  proportion  of  the  mixture  be  required,  the  same  proportion 
of  each  simple  must  evidently  be  taken.  The  same  would  be  done  if  3  times, 
6  times,  &c.,  one  simple  were  given. 

4.  A  merchant  has  two  kinds  of  sugar,  worth  8  cents  and 
13  cents ;  how  much  of  each  must  he  take  for  a  mixture  worth 
10  cents  per  pound  ? 

OPERATION. 

Price  of  mixture,  10  cents.  j  J]  |  j£  £  J  _ 

NOTE  4.  —  The  price  of  the  mixture  in  the  last  two  examples  is  the  same. 

5.  A  merchant  has  foul  kinds  of  sugar,  worth  6,  8,  13,  and 
17  cents  per  pound ;  how  much  of  each  must  he  take  for  a 
mixture  worth  10  cents  ? 


H197.                                   ALLIGATION.  249 

OPERATION.  SOLUTION. — Connecting 

7  6  with  17,  to  show  that 

o*  sugar  at  those  prices  are 


Price  of  mixture,  10  cents. 


to  be  mixed,  we  have  7 


2.     Ibs.  at  6  cents  and  4  Ibs. 
4.      at   17   cents,    forming  a 
mixture  of  11  Ibs.  worth 

10  cen's.     In  like  manner,  we  have  a  mixture  of  5  Ibs.  worth  10  cents, 
by  connecting  8  and  13.    And  11  -}-5  =  16  Ibs.  in  all. 

NOTE  5.  —  We  may  see  in  this  and  all  examples,  that  the  sum  of  the  defi 
ciencies  equals  the  sum  of  the  excesses. 

6.    A  merchant  has  3  kinds  of  sugar,  worth  6,  8,  and  15 
cents ;  how  much  of  each  must  he  take  for  a  mixture  worth 

1 1  cents  per  pound  ? 

OPERATION.  SOLUTION.  — 

/     g ,  4  We  have   not 

•n  •         f     •  j.          11  1     o_  two   pairs,   as 

Price  of  mixture,  11  cents.  ^         |    4.  •  in  $£  fo;mer 

(  15-^  5+3  =  8.     example,  but  4 
Ibs.  at  6  cents, 

may  be  mixed  with  5  Ibs.  at  15  cents,  and  4  Ibs.  at  8  cents,  with  3  Ibs. 
at  15  cents.  The  15  is  connected  with  both,  because  we  take  two  por 
tions  at  this  price,  5  Ibs.  to  mix  with  4  Ibs.  at  6  cents,  and  3  Ibs.  to  mix 
with  4  Ibs.  at  8  cents.  "We  have,  then,  8  Ibs.  at  15  cents. 

IT  197.     From  the  examples  explained,  We  have  the  fol 
lowing 

RULE. 

I.  Write   the  prices  of  the  simples  directly  under  each 
other,  beginning  with  the  least,  and  the  price  of  the  mixture 
at  the  left  hand. 

II.  Connect  each  price  less  than  the  mean  price  with  one 
or  more  greater,  and  each  price  greater  with  one  or  more  that 
is  less. 

III.  Write  the  difference  between  the  price  of  the  mixture 
and  of  each  simple  opposite  to  the  price  or  prices  with  which 
it  is  connected ;  the  number  or  numbers  opposite  to  the  price 
of  each  simple  will  express  its  quantity  in  the  mixture. 

IV.  If  any  fraction  or  multiple,  as  one  third,  or  three  times 
one  of  the  simples  is  to  be  taken,  the  same  fraction  or  multi- 
Questions.  —  IT  196.    Why  4  bushels  of  oats  to  1  of  corn,  Ex.  1  and  2? 

What  equalities  appear  from  Ex.  2  ?  How  could  we  have  made  such  a  mix 
ture  as  we  have,  if  the  prices  only  had  been  known?  Give  the  general  method 
of  forming  such  mixtures.  What  are  equal,  as  explained  in  note  2?  Why? 
What  must  be  done  if  a  greater  or  less  quantity  of  the  mixture  be  required?  — 
of  one  simple  ?  Why  are  numbers  connected  ?  What  is  done,  in  Ex.  6,  when 
.here  are  not  two  pairs  ?  What  is  alligation  alternate  ? 
IT  197.  What  is  the  rule  7 


250  ALLIGATION.  IT  197 

pie  of  each  of  the  other  simples  will  be  required.  Or,  if  any 
fraction  or  multiple  of  the  whole  compound  be  required,  the 
same  fraction  or  multiple  of  each  simple  must  be  taken. 

EXAMPLES    FOR    PRACTICE. 

1.  What  proportions  of  sugar,  at  8  cents,  10  cents,  and  14  cents 
per  pound,  will  compose  a  mixture  worth  12  cents  per  pound? 

Ans.  In  the  proportion  of  2  Ibs.  at  8  and  10  cents  to  6  Ibs.  at  14 
cents. 

2.  A  grocer  has  sugars,  worth  7  cents,  9  cents,  and  12  cents  per 
lb.,  which  he  would  mix  so  as  to  form  a  compound  worth  10  cents 
per  pound  ;  what  must  be  the  proportions  of  each  kind  ? 

Ans.   2  Ibs.  of  the  first  and  second,  to  4  Ibs.  of  the  third  kind. 

3.  If  he  use  1  lb.  of  the  first  kind,  how  much  must  he  take  of  the 

others?  if  4  Ibs.,  what?  if  G  Ibs.,  what?  if  10  Ibs., 

what?  if  20  Ibs.,  what? 

Ans.  to  the  last,  20  Ibs.  of  the  second,  and  40  of  the  third. 

4.  A  merchant  has  spices  at  16d.,  20d.,  and  32d.  per  pound;  he 
would  mix  5  pounds  of  the  first  sort  with  the  others,  so  as  to  form  a 
compound  worth  24d.  per  pound ;  how  much  of  each  sort  must  he 
use  ?  Ans.  5  Ibs.  of  the  second,  and  7£  Ibs.  of  the  third. 

5.  How  many  gallons  of  water,  of  no  value,  must  be  mixed  with 
SO  gallons  of  rum,  worth  80  cents  per  gallon,  to  reduce  its  value  to 
70  cents  per  gallon  ?  Ans.  8^-  gallons. 

6.  A  man  would  mix  4  bushels  of  wheat,  at  $1'50  per  bushel,  with 
rye  at  $1'16,  corn  at  $'75,  and  barley  at  $'50,  so  as  to  sell  the  mix 
ture  at  $'84  per  bushel ;  how  much  of  each  must  he  use? 

7.  A  goldsmith  would  mix  gold  17  carats  fine  with  some  19,  21, 
and  24  carats  fine,  so  that  the  compound  may  be  22  carats  fine ;  what 
proportions  of  each  must  he  use  ? 

Ans.  2  of  the  first  3  sorts  to  9  of  the  last. 

8.  If  he  use  1  oz.  of  the  first  kind,  how  much  must  he  use  of  the 
others  ?     What  would  be  the  quantity  of  the  compound  ? 

Ans.  to  the  last,  7£  ounces. 

9.  If  he  would  have  the  whole  compound  consist  of  15  oz.,  how 

much  must  he  use  of  each  kind  ?  if  of  30  oz.,  how  much  of  each 

kind?  if  of  37^  oz.,  how  much? 

Ans.  to  the  last,  5  oz.  of  the  first  3,  and  22i  oz.  of  the  last. 
10     A  man  would  mix  100  pounds  of  sugar,  some  at  8  cents,  some 
at  10  cents,  and  some  at  14  cents  per  pound,  so  that  the  compound 
may  be  worth  12  cents  per  pound ;  how  much  of  each  kind  must  he 
use 1-  20  Ibs.  at    8  cts.  ) 

20  Ibs.  at  10  cts.  V  Ans. 
60  Ibs.  at  14  cts.  ) 

11.  A  grocer  has  currants  at  4d.,  6d.,9d.,  and  lid.  per  lb.,and  he 
would  make  a  mixture  of  240  Ibs.,  so  that  the  mixture  may  be  sold 
at  8d.  per  lb. ;  how  many  pounds  of  each  sort  may  he  take? 

Ans.  72,  24,  48,  and  96  Ibs.,  or  48,  48,  72,  72,  &c. 

NOTE.  —  This  question  may  have  five  different  answers. 


IT  198,  199.  EXCHANGE.  251 


EXCHANGE. 

ff  198.  If  a  farmer,  A,  has  corn,  and  a  manufacturer,  B, 
has  cloth,  each  mo:e  than  he  needs  himself,  while  he  wants 
some  of  the  article  possessed  by  the  other,  an  exchange  will 
be  made  for  mutual  accommodation.  But  if  B  does  not  want 
A's  corn,  while  A  still  wants  the  cloth,  the  latter  must  find 
a  third  person,  if  possible,  who  wants  his  corn,  and  can  give 
him  something  for  it  which  B  may  want.  This  might  be 
difficult,  unless  some  article  was  settled  on,  which  every  one 
would  take  ;  then  A  might  exchange  his  corn  for  it,  as  B 
would  part  with  his  cloth  for  such  an  article,  since,  if  every 
one  would  take  it,  he  could  procure  with  it  whatever  he 
might  desire,  though  he  did  not  want  it  himself. 

Such  an  article  is  called  money.     Gold  and  silver,  contain 
ing  great  value  in  little  space,  are  employed  for  money  among 
civilized  nations,  and  sometimes  copper,  to  represent  smal' 
values.     This  exchange  between  individuals  is  called  trade, 
or  commerce. 

NOTE  1.  —  Since  gold  and  silver,  in  their  pure  state,  are  too  flexible  for  the 
purposes  of  a  circulating  medium,  nine  parts  of  pure  gold  and  one  part  of  sil 
ver  and  copper,  in  equal  quantities,  are  used  by  tne  U.  S.  government,  for  gold 
coins,  nine  parts  of  silver  and  one  of  copper,  for  silver  coins.  The  baser  metal, 
in  each  instance,  is  called  alloy.  The  English  government  uses  only  one  part 
of  alloy  to  eleven  of  the  gold,  and  a  little  less  alloy  in  silver  coins.  Hence, 
English  coins  are  more  valuable  than  ours  of  the  same  weight.  Again,  as 
coins  are  troublesome,  and  sometimes  expensive  to  transport,  and  also  suffer 
loss  by  wear,  bank  bills  are  much  used  for  circulation,  which,  though  valueless 
themselves,  are  readily  taken  as  money,  being  payable  in  specie,  on  demand, 
at  the  banks  which  issued  them.  The  coins  are  called  specie,  in  distinction 
from  paper  money,  and  together  they  form  what  is  called  the  circulating  me  • 
dium,  or  currency. 

NOTE  2.  —  By  the  fineness  of  gold,  is  meant  its  purity,  a  twenty-fourth  part 
of  any  quantity  being  called  a  carat.  When,  for  example,  there  are  two  parts 
of  alloy  to  twenty-two  of  pure  gold,  it  is  said  to  be  twenty-two  carats  fine. 

51"  199.  Bank  bills  can  be  used  in  trade  by  individuals 
of  the  same  country,  but  are  not  convenient  in  trade  between 
those  of  different  countries,  since  they  would  be  removed  too 
far  from  the  place  where  payable.  But  the  cost  and  risk  of 

Questions.  —  IT  198.  What  trade  are  A  and  B  supposed  to  make  ?  What 
will  A  do,  if  B  does  not  want  his  corn  ?  What  is  exchange ?  What  is  money? 
What  are  used  for  money,  and  why  ?  How  much  do  coins  want  of  being  pure 
gold  and  silver?  What  is  the  value  of  English  coins  compared  with  ours,  and 
why  ?  Why  are  bank  bills  used  ?  What  do  you  understand  to  be  the  differ 
ence  between  a  bank  bill  now  described,  and  a  bank  note,  TT  174?  What  is 
specie?  What  is  currency?  WThal  is  meant  by  fineness  of  gold?  —by 
carat  ?  —  by  carats  fine  ?  Illustrate. 


252  EXCHANGE.  1F  199. 

transporting  specie  5  considerable.  If,  for  instance,  Boston 
merchants  purchase  goods  in  Hamburgh  to  the  amount  of 
$2,050,000  a  year,  and  the  expense  of  transporting  specie  was 
3  per  cent.,  which  is  only  a  moderate  allowance,  this  expense 
would  be  $61,500  to  make  payments  for  one  year.  If,  on  the 
other  hand,  Hamburgh  merchants  should  purchase  $2,000.000 
worth  of  goods,  it  would  cost  them  $60,000  to  make  the  pay 
ments  in  specie. 

To  reduce  this  expense,  the  following  method  has  been  de 
vised.  A  B,  a  Boston  merchant,  has  $10,000  due  him  in 
Hamburgh,  from  C  D.  He  writes  an  order  for  the  sum,  and 
finds  some  one  who  owes  $10,000  to  another  merchant  in 
Hamburgh.  He  sells  to  him  this  order,  and  the  purchaser 
sends  it  to  his  creditor,  who  goes  to  C  D,  A  B's  debtor,  and 
receives  the  money.  In  this  way,  $2,000,000  of  the  Boston 
purchase  might  be  balanced  by  the  $2,000,000  purchased. 
in  Boston  by  Hamburgh  merchants,  and  the  Boston  mer 
chants  would  have  to  send  only  $50,000,  the  balance,  thus 
reducing  the  expense  of  making  payments  between  the  two 
ports  from  $121,500  to  $1500  ! 

Such  an  order  is  called  a  Bill  of  Exchange. 

NOTE  1 .  —  Lest  a  hill  of  exchange  may  be  lost,  or  delayed,  three  copies  are 
sent,  by  different  conveyances,  and  when  one  is  paid,  the  others  are  canceled. 
The  form  of  one  bill,  to  which  the  others  agree,  except  in  the  numbers  of  the 
bills,. is  here  given. 

Exchange  for  $10,000.  Boston,  Jan.  1,  1848. 

Three  months  after  date,  pay  this,  my  first  of  exchange, 
(second  and  third  of  the  same  tenor  and  date  not  paid,)  to  the 
order  of  F lemming  and  Johnson,  ten  thousand  dollars,  value 
received,  with,  or  without  further  advice  from  me.  A  B. 

C  D, 

Merchant  at  Hamburgh. 

NOTE  2.  If  A  B  had  to  be  at  the  expense  of  bringing  from  Hamburgh  the 
specie  on  his  due,  he  could  afford  to  sell  his  bill  for  less  than  $10,000,  on  ac 
count  of  freight.  This  he  would  have  to  do  when  more  was  to  be  paid  from 
Hamburgh  to  Boston  than  from  Boston  to  Hamburgh,  that  is,  if  the  balance 
of  trade  were  in  favor  of  Boston,  as  there  would  not  be  demand  for  all  the 

Questions.  —  IT  199.  Why  are  not  bank  bills  convenient  in  exchange 
oetween  different  countries?  What  difficulty  in  paying  with  specie?  Illus 
trate  by  the  example  of  trade  between  Boston  and  Hamburgh.  How  is  A  B 
supposed  to  get  his  pay  from  Hamburgh,  on  a  debt  of  $10,000?  How  much 
is  saved  by  this  means,  in  the  supposed  trade  between  the  two  ports  ?  Why 
must  $50,000  of  specie  be  sent  to  Hamburgh  ?  What  is  a  bill  of  exchange  ? 
What  care  is  taken  in  sending  bills?  Give  a  form.  What  would  be  the 
form  of  the  second  bill?  —  of  the  third  ?  When,  and  why,  is  exchange  belov 
«ar  ?  -  above  par  ?  Why  do  brokers  deal  in  exchange  ? 


IT  200.  EXCRA  NGE.  253 

orders  on  Hamburgh,  and  specie  wouVd  have  to  be  brought  on  some.  Ex 
change  is  then  said  to  be  below  par.  But  if  the  purchaser  had  to  pay  for 
freighting  the  money  he  owes  to  Hamburgh,  he  could  afford  to  pay  more  than 
Si 0,000  for  A  B's  bill,  which  he  would  have  to  do  when  the  balance  of  trad -i 
was  against  Boston,  that  is,  when  Boston  owed  more  to  Hamburgh  than  it  had 
owing,  for  then  there  would  not  be  orders  on  Hamburgh  sufficient  to  pay  the 
debts,  and  some  must  send  specie.  Exchange  is  then  said  to  be  above  pai ,  or 
at  a  premium. 

NOTE  3.  —  The  broker  is  the  medium  between  the  seller  and  purchaser  of 
bills,  since  the  former  might  not  know  to  whom  he  could  sell,  or  the  latter  of 
whom  he  could  buy.  Brokers  purchase  bills,  or  take  them  to  sell  on  commis 
sion. 

The  illustrations  of  the  principles  of  exchange  will  be  applied  to  exchange 
with  England  and  France. 


Exchange  with  England. 

IT  2OO.  The  nominal  value  of  the  pound  sterling  is 
$4'44|-,  consequently,  a  bill  of  exchange  for  £1000  is  said  to 
be  worth  $4444'44f .  But  by  comparing  the  materials  of  the 
English  sovereign,  a  gold  piece  representing  a  pound,  and  the 
eagle  of  our  currency,  the  former  is  worth  about  $4'86T6TT.  Sov 
ereigns,  however,  are  more  or  less  worn  by  use,  those  dated  as 
far  back  as  1821  being  worth  no  more  than  $4'80.  It  is  pre 
sumed  that  a  quantity  will  average  $4'84  each,  and  at  this 
value  they  are  taken  in  payment  of  duties.  If,  now,  the 
nominal  pound,  $4'44|,  be  multiplied  by  '09,  $4'44f  X  '09 
=  '40,  and  the  product  be  added  to  it,  $4'44f  -p40=$4'84$ 
it  will  be  changed  to  about  its  value,  in  the  custom-house  es 
timations.  If  '09J  be  added  to  the  nominal  pound,  it  will 
become  $4'86f,  nearly  its  commercial  value.  When,  then, 
sterling  exchange  is  quoted  at  9  or  9|  per  cent,  advance,  we 
must  understand  that  bills  sell  for  their  par  value.  When 
above  these  rates,  they  are  at  a  premium ;  when  below,  at  a 
discount. 

NOTE  1 .  —  In  the  following  examples  we  shall  consider  9£  per  cent,  above 
the  nominal,  the  par  value.  The  pupil  must  not  suppose,  however,  that  10  per 
cent,  above  the  nominal,  would  be  i  per  cent,  above  the  real  value  of  a  bill. 

1.  A  merchant  sells  a  bill  of  exchange  for  j£5000  at  its  par  value  ; 
what  does  he  receive?  Ans.  $24333'33£. 

2.  A  merchant  sold  a  bill  of  exchange  for  £7000  sterling,  at  11 
per  cent.  advan?e  ;  what  did  he  receive  more  than  its  real  value? 

Ans.  $466'66|. 

3.  A  merchant  sells  a  bill  on  Londen  for   £'4000  at  8  per  cent, 
above  its  nominal  value,  instead  of  importing  specie  at  an  expense  of  Si 
per  cent.  ;  what  does  he  save?  Ans.  $122'66f . 

22 


254  EXCHANGE.  H  201. 

4.  A  broker  sold  a  bill  of  exchange  for  jGSOOO,  on  commission,  at 
10  per  cent,  above  its  nominal  value,  receiving  a  commission  of  -^ 
per  cent,  on  the  real  value,  and  5  per  cent,  on  what  he  obtained  for 
the  bill  above  its  real  value  ;  wha ,  was  his  commission  ? 

Ans,  $ll'95f . 

NOTE  2.  —  Though  dollars  and  cents  are  the  denominations  of  U.  S.  money, 
shillings  and  pence  are  much  used  in  common  calculations.  But  the  dollar 
.ias  different  values  in  different  states,  as  expressed  in  shillings ;  thus,  in 
Now  York,  Ohio,  and  Michigan,  it  is  8  shillings  ;  in  North  Carolina,  10  shil 
lings  ;  in  New  Jersey,  Pennsylvania,  Delaware,  and  Maryland,  it  is  7  shillings 
6  pence  ;  in  Georgia  and  South  Carolina  it  is  4  shillings  8  pence,  while  in  the 
othc  i-  states  it  is  6  shillings.  The  change  of  monev  from  one  of  these  cur- 
reiu'ies  to  the  other  is  not  now  worthy  of  a  formal  discussion,  as  a  method 
will  readily  suggest  itself  to  the  practised  arithmetician,  and  the  custom  of 
using  these  denominations,  it  is  hoped,  will  be  speedily  given  up  for  the  sim 
pier  system  of  our  federal  currency.  Thus,  as  C  shillings  in  New  E«glan<? 
equal  8  shillings  in  New  York,  add  one  third  of  any  number  of  shillings  N.  E 
currency  to  the  number,  and  we  have  the  value  expressed  in  shillings  N.  Y 
currency . 


Exchange  with  France. 

^F  ££O1.  The  unit  of  French  money  is  the  franc,  the  value 
of  which  is  $'18f  .  In  the  quotations  of  French  exchange,  we 
have  the  number  of  francs  that  the  dollar  is  rated  at.  As  $VOO 

7 
is  equal  to  5^—^-  francs  ==5*3?^  -[">  when  a  dollar  is  worth 


francs  it  is  at  par. 

1.  A  New  York  merchant  sold  a  bill  of  exchange  for  $2500  on 
Havre,  at  5'4  francs  per  dollar  ;  what  did  he  obtain  for  it  more  than 
its  value?  Ans.  $11. 

2.  A  merchant  bought  a  bill  on  Havre  of  $2800  at  5'  31  francs  per 
dollar  ;  what  did  he  give  less  than  its  value?  Ans.  $34'  552. 

Questions.  —  IT  200.  What  is  the  nominal  value  of  the  English  pound  ? 
—  the  real  value?  What  are  sovereigns  of  1821  worth?  Why  no  more? 
What  is  the  average  value  of  sovereigns  supposed  to  be.  and  where  are  they 
taken  at  this  value?  How  is  the  pound  changed  from  its  nominal  to  its  real 
value  ?  What  is  added  to  the  nominal  value  in  the  examples  ?  What  is  said 
of  10  per  cent.  ?  What  denominations  are  still  used  in  common  calculations  ? 
What  are  the  different  values  of  the  dollar  in  different  states  ? 

IT  201.  What  is  the  unit  of  French  money?  —  its  value?  What  is  the 
par  value  of  the  dollar,  as  expressed  in  francs? 


11202. 


EXCHANGE. 


255 


IT  2OS.    Value  of  Gold  Coins. 

[According  to  the  Laws  passed  by  Congress,  May  and  June,  1834.] 


NAMES  OP  COINS. 

d.  c.  m. 

NAMES  OF  COINS. 

d.  c.  m. 

UNITED  STATES. 

HANOVER. 

Eaele,  coined  before  July  31, 

Double  George  d'or,  single  in 

1834, 

10  66  5 

proportion, 

7  87  9 

Shares  in  proportion. 

Ducat, 

2  29  6 

Gold  Florin,  double  in  propor 

FOREIGN  GOLD. 
AUSTRIAN  DOMINIONS. 

tion, 
HOLLAND. 

1  67  0 

Souverein, 

3  37  7 

Double  Ryder, 

12  20  5 

Double  Ducat, 

4  58  9 

Ryder, 

6  04  ? 

Hungarian,  do., 

2  29  6 

Ducat, 

2  27  f 

BAVARIA. 

Ten  Guilder  piece,  5  do.  in  pro 

Carol  in, 
Max  d'or,  or  Maximilian, 
Ducat, 

4  95  7 
3  31  8 
2  27  5 

portion, 
MALTA. 
Double  Louis, 

4  03  « 
9  27  b 

BERNE. 

Lou'.s, 

4  85  2 

Ducat,  double  in  proportion, 

1  98  6 

Demi  Louis, 

2  33  6 

Pistole, 

4  54  2 

MEXICO. 

BRAZIL. 

Doubloons,  shares  in  proportion," 

15  53  5 

Johannes,  £  in  proportion, 

17    6  4 

MILAN. 

Dobraon, 

32  70  6 

Sequin, 

2  29  0 

Dobra. 

17  30  1 

Doppia,  or  Pistole, 

3  80  7 

Moidore,  i  in  proportion, 

6  55  7 

Forty  Livre  Piece,  1808, 

7  74  2 

Crusade, 

635 

NAPLES. 

BRUNSWICK. 

Six  Ducat  Piece,  1783, 

5  24  9 

Pistole,  double  in  proportion, 

Ducat, 

4  54  8 
2  23  0 

Two  do.,  or  Sequin,  1762, 
Three  do.,  or  Oncetta,  1818, 

1  59  1 
2  49  0 

COLOGNE. 

NETHERLANDS. 

Ducat. 

2  26  7 

Gold  Lion,  or  Fourteen  Florin 

COLOMBIA. 

Piece. 

5  04  6 

Doubloons, 

15  535 

Ten  Florin  Piece,  1820, 

4  01  9 

DENMARK. 

PARMA. 

Ducat,  Current, 

1  81  2 

Quadruple    Pistole,    double    in 

Ducal..  Specie, 

2  26  7 

proportion, 

16  62  8 

Christian  d'or, 

4  02  1 

Pistole  or  Doppia,  1787, 

4  194 

EAST  INDIES. 

do.            do.,       1796, 

4  13  5 

Rupee.  Bombay,  J818, 

7  09  6 

Maria  Theresa,  1818, 

3  86  1 

Rupee:  Madras,  1818, 

7  11  0 

PIEDMONT. 

Pasoda.  Star, 

1  79  8 

Pistole,  coined  since  1785,  half 

ENGLAND. 

in  proportion, 

5  41  1 

Guinea,  half  in  proportion, 

5  07  5 

Sequin,  half  in  proportion. 

2  28  0 

Sovereign,  do., 

4  84  6 

Carlino.  coined  since  1785.  half 

Seven  Shilling  Piece, 

1  69  8 

in  proportion. 

27  34  0 

FRANCE. 

Piece  of  20  francs,   called  Ma 

Double    Louis,    coined    before 

ne  1120, 

3  56  4 

1786, 

9  69  7 

POLAND. 

Louis,  do., 

4  84  6 

Ducat. 

2  27  5 

Double  Louis,  coined  since  1786, 

9  15  3 

PORTUGAL. 

Louis,                do.            do., 

4  57  6 

Dobraon, 

32  70  6 

Double  Napoleon,  or  40  francs, 
Napoleon,  or  20  do., 
FRANKFORT  ON  THE  MAIN. 

7  70  2 
3  85  1 

Dobra, 
Johannes, 
MoMore,  half  in  p»oportion, 

17  30  1 
17  06  4 
6  55  7 

Ducat, 

227  9 

Piece  of  16  Testoons,  or  1600 

GENEVA 

Rees, 

2  12  1 

Pistole,  old, 

3  98  5 

Old  Crusado,  of  400  Rees, 

58  5 

Pistole,  new, 

3  44  4 

New  do.,  480  do., 

63  5 

GENOA. 

Milree.  coined  in  1775, 

78  0 

Sequin, 

2  30  2 

PRUSSIA. 

HAMBURG. 

Ducat,  1748, 

2  27  9 

Ducat,  double  in  proportion, 

327  9 

do.,    1787, 

2  26  7 

256 


DUODECIMALS. 


IT  203 


NAMES  OF  COINS. 

d.  c.  m. 

NAMES  OP  COINS. 

d.  c.m. 

Frederick,  double,  1769, 

7  95  5 

Coronilla,  Gold  Dollar,  or  Vin- 

do.            do.,     1800. 

7  95  1 

tern,  1S01, 

98  3 

do.         single,  1778, 

3  99  7 

SWEDEN. 

do.            do.,     1800. 

3  97  5 

Ducat. 

2  23  5 

ROME. 

SWITZERLAND. 

Sequin,  coined  since  1760, 
Scudo  of  Republic, 
RUSSIA. 

2  25  1 
15  81  1 

Pistole  of  Helvetic   Republic, 
1800, 
TREVES. 

56 

Ducat,  1796, 
do.,   1763, 

2  29  7 

2  26  7 

Ducat, 
TURKEY. 

226 

Gold  Ruble,  1756, 

96  7 

Sequin  Fonducli,  of  Constanti 

do.,            1799, 

73  7 

nople,  1773, 

1  86 

do.   Poltin,  1777, 

35  5 

do.,    1789, 

1  84 

Imperial,       1801, 

7  82  9 

Half  Misseir,  1818, 

52 

Half  do.,        1801, 

3  93  3 

Sequin-  Fonducli, 

1  83  : 

fc  iRDINIA. 

Yeermeeblekblek, 

3028 

Carli'io,  half  in  proportion, 
StXONY. 

9  47  2 

TUSCANY. 
Zechino,  or  Sequin, 

2  31  8 

Ducat,        1784, 

226  7 

Ruspone    of    the    kingdom   of 

do.,         1797, 

2  27  9 

Etruria, 

6  93  8 

Augustus,  1754, 

3  92  5 

VENICE. 

do.,          1784, 

3  97  4 

Zechino,  or  Sequin,  shares  in 

SICILY. 
Ounce.  1751, 

2  50  4 

proportion, 
W1RTEMBURG. 

2  31  0 

Double  do.,  1758, 

5  044 

Carolin, 

4  89  8 

SPAIN.' 

Ducat. 

2235 

Doubloon.  1772.  double  and  sin 

ZURICH. 

gle,  and  shares  in  proportion, 

16  02  3 

Ducat,  double  and  half  in  pro 

Doubloon, 

15  53  5 

portion, 

2  26  7 

Pistole, 

3  88  4 

DUODECIMALS. 

5F  9O3.  Duodecimals  are  fractions  of  a  foot.  The  word 
is  derived  from  the  Latin  word  duodecim,  which  signifies 
twelve.  A  foot,  instead  of  being"  divided  decimally  into  ten 
equal  parts,  is  divided  duodecimally  into  twelve  equal  parts, 
called  primes,  marked  thus  (').  Again,  each  of  these  parts 
'.s  conceived  to  be  divided  into  twelve  other  equal  parts, called 
seconds,  (")•  In  like  manner,  each  second  is  conceived  to 
be  divided  into  twelve  equal  parts,  called  thirds  ("'}',  each 
third  into  twelve  equal  parts,  called  fourths  ("");  and  so  on 
to  any  extent. 

In  this  way  of  dividing  a  foot,  it  is  obvious,  that 


1'  prime  is     .       :    .     .    -.     .     .  . 

1"  second  is  T^-  of  fa,  .     .     .     .  = 

V"  third  is  -fa  of  -fa  of  fa,      .     .  = 

1"  fourth  is  fa  of  ~fa  of  -fa  of  -fa,  = 

1"  fifth  is-- 


of  a  foot. 
of  a  foot. 
of  a  foot. 
of  a  foot. 

°f  a  f°ot» 


IT  204.  DUODECIMALS.  257 

TABLE. 

12""fft-jrths  make  I'"  third, 
12"'  thirds  .     .     .  1"  second, 
12"    seconds    .     .1'    prime, 
12'     primes,     .     .  1     foot. 

NOTE  1.  — The  marks, ',  ",  '",  "",  &c.,  which  distinguish  the  different  jarts, 
are  called  the  indices  of  the  parts  or  denominations. 

NOTE  2.  —  The  divisions  of  a  unit  in  duodecimals  are  uniform,  just  as  in 
decimal  fractions,  with  this  difference  :  they  decrease  in  a  hcelve-fold  propor 
tion,  12  of  a  lower  denomination  making  t  of  a  higher.  Operations  in  Inern 
are  consequently  the  same  as  in  whole  numbers  or  decimals,  except  that  12  is 
the  carrying  number  instead  of  10. 


Multiplication  of  Duodecimals. 

^T  2O4.  Duodecimals  are  used  in  measuring  surfaces  and 
solids. 

1.  How  many  square  feet  in  a  board  16  feet  7  inches  long, 
and  1  foot  3  inches  wide  ? 

NOTE.  — Length  X  breadth  s=  superficial  contents,  (IT  48.) 

OPERATION.  SOLUTION.  —  7  inches,  or  primes,  =  fa  of 

/*•  a  foot,  and  3  inches  =  -$%•  of  a  foot ;  conse- 

Length,    16     7  quently.  the  product  of  7'  X  3'  =  •£&  of  a 

breadth,     1     3  foot,  that  is,  21"  =  1'  and  9";  wherefore  we 

set  down  the  9",  and  reserve  the  I7  to  be  car 

4      ]/     9"      ried  forward  to  its  proper  place.     To  multi- 
\Q     7'  ply  16  feet  by  3',  is  to  take  f^-  of  JT6-  =  f  f, 

that  is,  48';  and  the  I'  which  we  reserved 

A        on     e'     Q"      makes  49'.  =  4  feet  1' ;  we  therefore  set  down 
Ans.  ^0  the  i'5  an(j  carrv  forward  the  4  feet  to  its 

proper  place.  Then,  multiplying  the  multi 
plicand  by  the  1  foot  in  the  multiplier,  and  adding  the  two  products  to 
gether,  we  obtain  the  Answer,  20  feet  8'  and  9". 

NOTE  1.  —  In  all  cases  the  product  of  any  two  denominations  loill  always  be, 
of  the  denomination  denoted  by  the  sum  of  their  IKDICES.  Thus,  in  the  above 
example,  the  sum  of  the  indices  of  T'Xa'is  " ;  consequently,  the  product  is 
21':  and  thus  primes  multiplied  by  primes,  produce  seconds;  primes  multi 
plied  by  seconds,  produce  thirds ;  fourths  multiplied  byjifths,  produce  ninths, 

Questions.  —  IF  203.  What  are  duodecimals  ?  Explain  the  duodecimal 
divisions  and  subdivisions  of  a  foot.  Repeat  the  table.  What  are  indices? 
What  part  of  a  foot  is  1'?  —  1"?  —  \"'\—  1""?  —  \'""1  What  difference 
between  the  decimal  and  duodecimal  divisions  of  a  unit  ?  How  are  operations 
on  duodecimals  performed  ? 

22* 


258  DUODE:IMALS.  1F204 

2.    How  many  solid  feet  .n  a  block  15  ft.  8'  long,  1  ft.  5' 
wide,  and  1  ft.  4'  thick  ? 

OPERATION. 

ft- 

Length,       15     8' 
Breadth,        1     5' 


22     2'     4" 
Thickness,     1     4' 


6     6'     4" 

5      g'  The  length  multiplied  by  the  breadth. 

•  and  that  product  by  the  thickness,  gives 

the  solid  contents,  (IT  51.) 


7     4'     9"    4" 
22    2'     4" 

.  29     7'     1"    4'" 

Hence,  To  multiply  duodecimals, 

RULE. 

I.  Write  the   multiplier  under  the  multiplicand,  like  de 
nominations  under  like,  and  in  multiplying,  remember  that 
the  product  of  any  two  denominations  will  be  of  that  denomi 
nation  denoted  by  the  sum  of  their  indices. 

II.  Add  the  several  products  together,  and  their  sum  will 
be  the  product  required. 

EXAMPLES    FOR    PRACTICE. 

3.  How  many  square  feet  in  a  stock  of  15  boards,  each  of  which  is 
/2  ft.  8'  in  length,  and  13'  wide?  Ans.  205  ft.  10'. 

4.  What  is  the  product  of  371  ft.  2'  6"  multiplied  by  181  ft.  1'  9"? 

Ans.  67242  ft.  10'  1"  4'"  6"". 

5.  There  is  a  room  plastered,  the  compass  of  which  is  47  ft.  3', 
and  the  hight  7  ft   6' ;  what  are  the  contents  ] 

Ans.  39  yds.  3  ft.  4'  G". 

6.  What  will  it  cost  to  pave  a  court  yard,  26  ft."  8'  long  by  24  ft. 
9  wide,  at  $'90  per  square  yard?  Ans.  $66. 

7.  There  is  a  house  containing  two  rooms,  each  16  ft.  by  15  ft.  4'; 
a  hall  24  ft.  by  10  ft.  6';  three  bed-rooms,  each  11  ft.  4'  by  8  ft. ;  a 
pantry  7  ft.  by  9  ft.  6'  ;  a  kitchen  14  ft.  2'  by  18  ft.,  and  two  cham- 

Questioiis.  —  tf  20 1.  For  what  are  duodecimals  used  ?  Of  what  de 
nomination  is  the  product  of  any  two  denominations?  Repeat  the  rule  for  the 
multiplication  of  duodecimals.  *  How  do  you  carrv  frrm  one  denomination  to 
another?  How  is  masons' work  estimated?  What  is  understood  by  girt, 
an*l  for  what  used  ? 

1 


U  205.  DUODECIMALS.  259 

bers,  each  16  ft.  by  2C  \.  8' ;  what  did  the  work  of  flooring  cost,  at 
$'02  per  square  foot?  Ans.  $39'95. 

NOTE  2.  —Masons'  wjrk  is  estimated  bv  the  perch  of  16£  feet  in  length,  1J 
feet  in  width,  and  1  foot  in  hight.  A  perch  contains  24 '75  cubic  feet.  If  any 
wall  be  1*  feet  thick,  its  contents  in  perches  may  be  found  by  dividing  its  su 
perficial  contents  by  16*  ;  but  if  it  be  any  other  thickness  than  H  feet,  its  cubic 
contents  must  be  divided  by  24'75,  (=243,)  to  reduce  it  to  perches. 

Joiners,  painters,  plasterers,  brick-layers,  and  masons,  make  no  allowance 
for  windows,  doors,  &c.  Brick-layers  and  masons  make  no  allowance  for 
corners  to  the  walls  of  houses,  cellars,  &c.,  but  estimate  their  work  by  the 
girt,  that  is,  the  length  of  the  wall  on  the  outside. 

8.  The  side  walls  of  a  cellar  are  each  32  ft.  6'  long-,  the  end  walls 
24  ft.  6',  and  the  whole  are  7  ft.  high,  and  1£  ft.  thick  ;  how  many 
perches  of  stone  are  required,  allowing  nothing  for  waste,  and  for  how 
many  must  the  mason  be  paid  1 

.         (  45^T  perches  in  the  wall. 

(  The  mason  must  be  paid  for  48^T  perches. 

9.  How  many  cord  feet  of  wood  in  a  load  7  feet  long,  3  feet  wide, 
and  3  feet  4  inches  high,  and  what  will  it  cost  at  $'40  per  cord  foot? 

Ans.  4f  cord  feet,  and  it  will  cost  $1'75. 

10.  How  much  wood  in  a  load  10  ft.  in  length,  3  ft.  9'  in  width, 
and  4  ft.  8'  in  hight?  and  what  will  it  cost  at  $1'92  per  cord? 

Ans.  I  cord  and  2||  cord  feet,  and  it  will  cost  $2'62j. 

^T  SO5.  By  some  surveyors  of  wood,  dimensions  are 
taken  in  feet  and  decimals  of  a  foot.  For  this  purpose,  make 
a  rule  or  scale  4  feet  long,  and  divide  it  into  feet,  and  each 
foot  into  ten  equal  parts.  Such  a  rule  will  be  found  very 
convenient  for  surveyors  of  wood  and  of  lumber,  for  painters, 
joiners,  &c. ;  for  the  dimensions  taken  by  it  being  in  feet  and 
decimals  of  a  foot,  the  casts  will  be  no  other  than  so  many 
operations  in  decimal  fractions. 

1.  How  many  square  feet  in  a  hearth  stone,  which,  by  a  rule,  as 
above  described,  measures  4'5  feet  in  length,  and  2'6  feet  in  width ? 
and  what  will  be  its  cost,  at  75  cents  per  square  foot? 

4ns.  11'7  feet ;  and  it  will  cost  $8'775. 

2.  How  many  cords  in  a  load  of  wood,  7'5  feet  in  length,  3'6  feet 
in  width,  and  4'8  in  hight?  Ans.  1  cord  l-j-6^  en.  ft. 

3.  How  many  cord  feet  in  a  load  of  wood  10  feet  long,  3'4  feet 
wide,  and  3'5  high?  Ans.  7T7F. 

Questions.  — r.  205.  How  do  some  surveyors  of  wood  take  dimen- 
s  ons  7  Explain  th<j  rule  used  in  measuring.  How  are  dimensions  taken  by 
i.  estimated  7 


260 


INVOLUTION. 


11206 


Fourth  power. 


INVOLUTION. 

First poiver.     IT  2O6.     Three  feet  in  length  (IF  111)  are 

a  yard,  linear  measure ;  3  in  length  and  3  in 

width,   3x3  =  9   square    feet,   are   a  yard, 
Second  power,  square  measure;  3  in  length,  3  in  width,  and 
3  in  Light,  3  X  3  X  3  =  27  solid  feet,  are  a 
yard,  cubic  measure,  (1F113.) 

When  a  number,  as  3,  is  multiplied  into 
itself,  and  the  product  by  the  original  number,  and  so  on,  the 
series  of  numbers  produced  are  called  pmvers,  and  the  process 
of  producing  them  is  called  Involution. 

rpi .   ,  The  first  number,  represented  by  a  line,  is 

'  called  the  first  power,  or  root ;  the  second,  rep 
resented  by  a  square,  is  called  the  square,  or 
2d  power ;  the  third,  represented  by  a  cube,  is 
called  the  cube,  or  .3d  power. 

The  4th  power  of  3  is  3  times  the  3d 
power,  3  blocks   like  that  employed  to 
represent  the  3d  power,  and  may  be  rep 
resented  by  a  figure  3  times  as  large,  that 
is,  3  feet  wide,  3  feet  high,  and  9  feet  long. 
The  5th  power,  by  3  times  such 
a  figure,  or  one  3  high,  9  wide,  and 
9  long. 

The  6th  power,  3  times  this,  by 
a  figure  9  long,  9  wide,  and  9 
high,  or  a  cube. 

Thus  it  may  be  shown  that  the 
9th,  12th,  15th,  18th,  &c.,  powers, 
may  be  represented  by  cubes  ;  the 
7th,  10th,  13th,  16th,  &c.,  by  fig 
ures  having  greater  length  than 
width  and  hight;  the  8th,  llth, 
14th,  17th,  &c.,  by  figures  having 
greater  length  and  width  than 
hight. 

To  involve  a  number,  take  it 


Fifth  power. 


Sixth  power. 


as 


a  factor  as  many  times  as  is  indicated  by  the  required  power. 


NOTE.  1  —  The  number  denoting  the  pwer  is  called  the  index,  or  expo 
nent;  thus,  6*  denotes  that  5  is  raised  or  involved  to  the  4th  power. 


U206. 


INVOL  JTION. 


261 


EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  square,  or  2d  power,  of  71  Ans.  49. 

2.  What  is  the  square  of  30  ?  Ans.  900. 

3.  What  is  the  square  of  4000  ?  Ans.  16000000. 

4.  What  is  the  cube,  or  3d  power,  of  4  ?  Ans.  64. 

5.  What  is  the  cube  of  800  ?  Ans.  512000000. 

6.  What  is  the  4th  power  of  60  1  Ans.  12960000. 

7.  What  is  the  square  of  1?  -  of  2?  -  of  3?  -  of  4? 

Ans.  1,  4,  9,  and  16. 

8.  What  is  the  cube  of  1  ?  -  of  2?  -  of  3  ?  -  of  4? 

Ans.  1,  8,  27,  and  64. 

9.  What  is  the  square  of  f  ?  -  of  f  ?  -  of  £? 

Ans.  |,  £f  ,  and  f  f  . 

10.  What  is  the  cube  of  §  1  -  of  f  ?  -  of  £? 

>,  and  fff. 


11.  What  is  the  square  of  &  1  -  the  5th  power  of  £  ? 

Ans.  J  and  -g^. 

12.  What  is  the  square  of  1<5  ?  -  the  cube? 

Ans.  2'25,  and  3'375. 

13.  What  is  the  6th  power  of  1<2  ?  Ans.  2'985984. 

14.  Involvs  2|  to  the  4th  power. 

NOTE  2.  —  A  mixed  number,  like  the  above,  may  be  reduced  to  an  im 
proper  fraction  before  involving  ;  thus,  2i  =  %  ;  or  it  may  be  reduced  to  a 
decimal  ;  thus,  2i  =3  2'25. 


15.  What  is  the  square  of  4f-  ?  Ans.  -if  f  ^  =  23f  £  . 

16.  What  is  the  value  of  74,  that  is,  the  4th  power  of  7  ? 

Ans.  2401. 

17.  How  much  is  93?  -  6s?  -  104? 

Ans.  729,  7776,  10000. 

18.  How  much  is  2r?  -  36?  -  4s?  -  53?  -  65?  - 
108  ?  Ans.  to  the  last,  100000000. 

NOTE  3.  —  The  powers  of  the  nine  digits,  from  the  first  power  to  the  fifth 
may  be  seen  in  the  flowing 

TABLE. 


Roots   .     . 

|  or 

1st  Powers 

i|    2| 

5] 

4 

5 

6| 

7| 

8|          9 

Squares     . 

(or 

2d   Powers 

1  |    4| 

9| 

16  |      25 

36  | 

49  | 

64 

81 

Cubes  .     . 

|or 

3d  Powers 

1  |    8| 

27  | 

64 

125  |    216  | 

343) 

512 

729 

Biquadrates  |  or  4th  Powers 

1  |  10  | 

81  | 

256 

625  |  1296  | 

2401  | 

4096  |    6;"  61 

Sursolids  . 

|or 

5tli  Powers 

1  |  32  | 

243  | 

1024  |  3125 

7776  | 

16807  | 

32763 

59049 

Questions.  —  IT  206.  What  are  powers?  How  is  the  first  power  rep 
resented?  Why  is  the  second  power  called  the  square?  Why  the  third 
called  the  cube  ?  How  is  the  fourth  power  represented  ?  —  the  fifth  ?  —  the 
sixth  ?  —  thirtenth  ?  —  the  fourteenth  ?  —  the  twenty-first  ?  —  the  twenty- 
third  ?  —  the  twenty-fifth ?  What  is  involution ?  Hrw  is  a  number  involved 
to  any  power  ?  What  is  the  index,  and  how  written  ?  How  is  a  mixed  num 
ber  involved? 


EVOLUTION.  H  207,  208. 


EVOLUTION. 

^T  2OT.  Evolution,  or  the  extracting  of  roots,  is  the 
method  of  finding  the  root  of  any  power  or  number. 

The  root,  as  we  have  seen,  is  that  number,  which,  by  a 
continual  multiplication  into  itself,  produces  the  given  power, 
and  to  find  the  square  root  of  a  number  (one  side  of  a  square 
when  the  contents  are  given)  is  to  find  a  number,  which, 
being  squared,  will  produce  the  given  number ;  to  find  the 
cube  root  of  a  number  (the  length  of  one  side  of  a  cubic  body 
when  the  solid  contents  are  given)  is  to  find  a  number,  which, 
being  cubed  or  involved  to  the  3d  power,  will  produce  the 
given  number:  thus,  the  square  root  of  144  is  12,  because 
122=  144;  and  the  cube  root  of  343  is  7,  because  73,  that  is, 
7  X  7  X  7  =  343  ;  and  so  of  other  numbers. 

NOTE.  —  Although  there  is  no  number  which  will  not  produce  a  perfect 
power  by  involution,  yet  there  are  many  numbers  of  which  precise  roots  can 
never  be  obtained.  But,  by  the  help  ot  decimals,  we  can  approximate,  or  ap 
proach,  towards  the  root  to  any  assigned  degree  of  exactness.  Numbers, 
whose  precise  roots  cannot  be  obtained,  are  called  surd  numbers,  and  those 
whose  roots  can  be  exactly  obtained  are  called  rational  numbers. 

Tlie  square  root  is  indicated  by  this  character  ^/  placed  before  the  num 
ber  ;  the  other  roots  by  the  same  character,  with  the  index  of  the  root  placed 
over  it.  Thus,  the  square  root  of  16  is  expressed  y^/16  ;  and  the  cube  root  of 
27  is  expressed  £f  27  ;  and  the  5th  root  of  7776,  /%/77&7. 

When  the  power  is  expressed  by  several  numbers,  with  the  sign  -f-  or  — 
between  them,  a  line,  or  vinculum,  is  drawn  from  the  top  of  the  sign  over  all 
the  parts  of  it ;  thus,  the  square  root  of  21  — 5  is  y\/21 — 5. 


Extraction  of  the  Square  Root. 

IT  2O8.  1.  Supposing  a  man  has  625  yards  of  carpet 
ing,  a  yard  wide,  what  is  the  length  of  one  side  of  a  square 
room,  the  floor  of  which  the  carpeting  will  cover?  that  is, 
what  is  one  side  of  a  square,  which  contains  625  square 
yards  ? 

SOLUTION.  —  We  may  find  one  side  of  a  square  containing  625  square 
yards,  that  is,  the  square  root  of  625,  by  a  sort  of  trial ;  and, 

Questions.  —  IT  2O7.  What  is  evolution?  What  is  a  root?  —  tne 
square  root,  and  how  found  ?  —  the  cube  root,  and  how  found  ?  Give  exam 
ples.  WThat  do  you  say  of  perfect  powers  and  perfect  roots?  Give  the  dis 
tinction  between  surd  and  rational  numbers.  How  is  the  square  root  indi- 
cated  7  —  the  cube  root  ?  Describe  the  manner  of  using  the  vinculum. 


1T208. 


EVOLUTION. 


263 


20 
20 


400 


1st.  We  wiB.  endeavor  to  ascertain  how  many  figures  there  will  be  in 
the  root.  This  Tve  can  easily  do,  by  pointing  off  the  number,  from  units, 
into  periods  of  two  figures  each  ;  for  the  square  of  and  root  always  con 
tains  just  twice  as  many,  or  one  figure  less  than  twice  as  many  figures, 
as  are  in  the  root.  The  square  of  3  (3  X  3  =  9)  contains  1  figure ,  the 
square  of  4  (4  X  4  =  16)  contains  2  figures  ;  the  square  of  9  (9  X  9  = 
1)  contains  2  figures  ;  the  square  of  10 

OPERATION  (10  X  10  =  100)  contains  3  figures ;  the 

.    .  square  of  32  (32  X  32  =  1024)  contains 

625  (2  4  figures  ;  the  square  of  99  (99  X  ^  = 

9801)  contains  4  figures;  the  square  of 

100  (100  X  100  =  10000)  contains  5  fig- 

~  ures,  and  so  of  any  number.     Pointing 

225  off  the  number,  we  find  that  the  root 

will  consist  of  two  figures,  a  ten  and  a 

FIG.  I.  unit- 

,  2d.   We  will  now  seek  for  the  first  fig 

ure,  that  is,  for  the  tens  of  the  root,  which 
we  must  extract  from  the  left  hand  peri 
od,  6,  (hundreds.)  The  greatest  square 
in  6  (hundreds)  we  find  to  be  4,  (hun 
dreds,)  the  root  of  which  is  2,  (tens,  = 

2Q  20 ;  therefore,  we  set  2  (tens;  in  the  root. 

Since  the  root,  is  one  side  of  a  square,  let 
us  form  a  square,  (A,  Fig.  I.,)  each  side 
of  which  shall  be  regarded  2  tens,  =  20 
yards  long. 

The  contents  of  this  square  are  20  X 

u  20  o      20  =  400  yards,  now  disposed  of,   and 

which,  consequently,  are  to  be  deducted 

from  the  whole  number  of  yards,  (625,)  leaving  225  yards.  This  de 
duction  is  most  readily  performed  by  subtracting  the  square  number,  4, 
(hundreds,)  or  the  square  of  2,  (tens,)  from  the  period  0,  (hundreds.)  and 
bringing  down  the  next  period  to  the  remainder,  making  225. 

3d.  The  square  A  is  now  to  be  enlarged  by  the  addition  of  the  225 
remaining  yards ;  and  in  order  that  the  figure  may  retain  its  square 
form,  the  addition  must  be  made  on  two  sides.  Now,  if  the  225  yards 
be  divided  by  the  length  of  the  two  sides,  (20-4-20  =  40,)  the  quotient 
will  be  the  breadth  of  this  new  addition  of  225  yards  to  the  sides  c  d  and 
b  c  of  the  square  A. 

But  our  root  already  found,  =  2  tens,  is  the  length  of  one  side  of  the 
figure  A ;  we  therefore  take  double  this  root,  =  4  tens,  for  a  divisor. 

The  divisor,  4,  (tens,)  is  in  reality  40,  and  we  are  to  seek  how  many 
times  40  is  contained  in  225,  or,  which  is  the  same  thing,  we  may  seek 
how  many  times  4  (tens)  is  contained  in  22,  (tens,)  rejecting  the  right 
hand  figure  of  the  dividend,  because  we  have  rejected  the  cipher  in  the 

Questions.  —  IT  208.  How  may  one  side  of  a  square,  when  the  contents 
are  given,  be  found  ?  Why,  in  the  trial,  must  we  point  off  the  number  into 
periods  of  two  figures  each  ?  Illustrate.  Why  is  the  first  root  figure  2  tens  ? 
Of  what  number1  is  2  tens  the  root?  What  may  now  be  formed?  How  large? 
How  must  it  b«  increased  ?  What  will  be  the  divisor?  —  the  dividend?  — 
the  quotient?  Why  is  the  divisor  too  small?  What  is  the  entire  divisor? 
How  are  the  contents  of  the  addition  found  ?  What  does  Fig.  II.  represent  ? 
What  is  the  first  method  of  proof?  —  the  second  method? 


264 


EVOLUTION. 


IT  209. 


OPERATION  -  CONTINUED. 

625(25 
4 

45)225 
225 


FIG.  II. 

20  yds. 


5  yds. 


20 

B 

100 

5 

D  J 

25 

d                            c 
A 

C 

20 
20 

20 
5 

400 

100 

a                             I 

20  yds. 


5yds. 


The  proof  may  be  seen  by  adding 


proo    m 
,  thus  :  — 


divisor.  We  find  our  quotient, 
that  is,  the  breadth  of  the  addi 
tion  to  be  5  yards ;  but,  if  we 
look  at  Fig.  II.,  we  shall  perceive 
that  this  addition  of  5  yards  to 
the  two  sides  does  not  complete 
the  square  ;  for  there  is  still 
wanting,  in  the  corner  D,  a  small 
square,  each  side  of  which  is 
equal  to  this  last  quotient,  5  ;  we 
must,  therefore,  add  this  quotient, 
5,  to  the  divisor,  40,  that  is,  place 
it  at  the  right  hand  of  the  4, 
(tens.)  making  the  length  of  the 
whole  addition  formed  by  225 
square  yards,  45  yards  ;  and  then 
the  whole  length  of  the  addition, 
45  yards,  multiplied  by  5,  the 
number  of  yards  in  the  width,  will 
give  the  contents  of  the  whole 
addition  around  the  sides  of  the 
figure  A,  which,  in  this  case,  be 
ing  225  yards,  the  same  as  our 
dividend,  we  have  no  remainder, 
and  the  work  is  done.  Conse 
quently,  Fig.  II.  represents  the 
floor  of  a  square  room,  25  yards 
on  a  side,  which  625  square  yards 
of  carpeting  will  exactly  cover, 
together  the  several  parts  of  the 

Or  we  may  prove  it  by  involu 
tion,  thus  :  —  25  X  25  =  625,  as 
before. 


figure 

The  square  A  contains  400  yards. 
The  figure  B        «        100      " 

«        «      C        "        100      " 

«        «      D        "         25      " 

Proof,  625      « 

IF  2O9.     From  this  example  and,  illustration  we  derive  the 
folloiving  general 

RUDE 

FOR  THE  EXTRACTION  OF  THE  SQUARE  ROOT. 

I.  Point  off  the  given  number  into  periods  of  two  figures 
each,  by  putting  a  dot  over  the  units,  another  over  the  hun 
dreds,  and  so  on.     These  dots  show  the  number  of  figures  of 
which  the  root  will  consist. 

II.  Find  the  greatest  square  number  in  the  left  hand  pe 
riod,  and  write  its  root  as  a  quotient  in  division.    Subtract  the 
square  number  from  the  left  hand  period,  and  to  the  remain 
der  bring  down  the  next  period  for  a  dividend. 


If  209. 


EVOLUTION. 


265 


III.  Double  the  root  already  found  for  a  divisor ;  seeK  how 
many  times  the  divisor  is  contained  in  the  dividend,  excepting 
the  right  hand  figure,  and  place  the  result  in  the  root,  and 
also  at  the  right  hand  of  the  divisor;  the  divisor  thus  in 
creased  will  be  the  length  of  the  whole  addition  now  made  to 
two  sides  of  the  square  ;  multiply  the  divisor,  or  length  of  the 
addition,  by  the  last  figure  of  the  root,  (the  breadth  of  the 
addition,)  and  subtract  the  contents  of  the  addition  thus  ob 
tained  from  the  dividend,  and  to  the  remainder  bring  down 
the  next  period  for  a  new  dividend. 

IV.  Double  the  root  already  found  for  a  new  divisor,  and 
continue  the  operation  as  before,  until  all  the  periods  are 
brought  down. 

NOTE  1.  — As  the  value  of  figures,  whether  integers  or  decimals,  is  deter 
mined  by  their  distance  from  the  place  of  units,  we  must  always  begin  at 
units'  place  to  point  off  the  given  number,  and  if  it  be  a  mixed  number,  we 
must  point  it  off  both  ways  from  units,  and  if  there  be  but  one  figure  in  any 
period  of  decimals,  a  cipher  must  be  added  to  it.  And  as  the  root  must 
always  consist  of  as  many  integers  and  decimals  as  there  are  periods  belong 
ing  to  each  in  the  given  number,  when  it  is  necessary  to  carry  the  operation  to 
a  greater  degree  of  exactness  by  decimals  in  the  root,  after  all  the  periods  are 
brought  down,  two  ciphers,  a  whole  period,  must  be  annexed  for  every  decimal 
figure  which  we  would  obtain  in  the  root. 

EXAMPLES    FOR    PRACTICE. 

2.    What  is  the  square  root  of  61504  ? 


ILLUSTRATION. 


,240X8  =  1920                   64 

200X40  =  8000 

40 
40 
1600 

(NO 
0 

X 

00 

II 

1—  ' 
to 
to 

o 

200 
200 

200X40  =  8000 

40000 

OPERATION. 

61504(248  Ans 
4 


44)215 
176 

488)3904 
3904 


"200  4-40+ B 

PROOF.  — 40000  4-  8000  -f-  8000  -f  1600  4-  1920 
61504,  or  248  X  248  =  61504. 


0000 

The  pupil  will  easily  illustrate  the 
operation  by  the  annexed  diagram. 


1920  4-  64 


Questions.  —  IT  209.  What  is  the  general  rule  for  extracting  the  square 
root'?  _  Where  must  we  begin  to  point  off?  What  is  done  when  one  decimal 
place  is  wanting  ?  How  is  the  operation  continued,  when  all  the  periods  are 
brought  down?  Why  cannot  the  precise  root  be  found  when  tKere  is  a  re 
mainder  ?  How  is  the  root  of  a  fraction  obtained  ?  Why  ?  What  is  done 
when  the  terms  of  the  fraction  are  not  exact  squares  ? 
23 


266  EVOLUTION.  If  210. 

3.  What  is  the  square  root  of  43264  ?  Ans.  208. 

4.  What  is  the  square  root  of  998001  ?  Ans.  999. 

5.  What  is  the  square  root  of  234'09  ?  Ans.  15'3. 

6.  What  is  the  square  root  of  964'5192360241  ? 

Ans.  31*05671. 

7.  What  is  the  square  root  of  '001296?  Ans.  '036. 

8.  What  is  the  square  root  of  '2916  ?  Ans.  '54. 

9.  What  is  the  square  root  of  36372961  ?  Ans.  6031. 

10.  What  is  the  square  root  of  164?  Ans.  12'8-}~. 

NOTE  2.  —  In  the  last  example,  there  was  a  remainder,  after  a...  the  figures 
were  brought  down.  In  such  cases,  the  precise  root  can  never  be  obtained. 
For,  as  the  operation  is  continued  by  annexing  ciphers,  the  last  figure  of  every 
dividend  must  be  a  cipher.  But  the  root  figure  obtained  from  this  dividend, 
is  also  placed  at  the  right  hand  of  the  divisor,  and  consequently  is  multiplied 
into  itself,  and  the  last-  figure  of  the  product  placed  under  the  cipher,  which  is 
the  last  figure  of  the  dividend,  to  be  subtracted  from  it.  And  as  the  product 
of  no  one  of  the  significant  figures  ends  in  a  cipher,  there  will  always  be  a  re 
mainder. 

11.  What  is  the  square  root  of  3  ?  Ans.  1'73  4-. 

12.  What  is  the  square  root  of  10?  Ans.  3'  16-)-. 

13.  What  is  the  square  root  of  184'2?  Ans.  13'57-j-. 

14.  What  is  the  square  root  of  |-? 

NOTE  3.  —  Since,  from  the  rule  for  multiplying  one  fraction  by  another,  a 
fraction  is  involved  by  involving  its  numerator  and  its  denominator,  the  root 
cf  a  fraction  is  obtained  by  finding  the  root  of  its  numerator,  and  of  its  denom- 

:nator-  Ans.     . 


15.  What  is  the  square  root  of  /¥?  Ans.  f. 

16.  What  is  the  square  root  of  y1^?  Ans.  -£$. 

17.  What  is  the  square  root  of  T8¥L?  Ans.  -&=  f. 

18.  What  is  the  square  root  of  20£  ?  Ans.  4£. 

NOTE  4.  —  When  the  numerator  and  denominator  are  not  exact  squares^  the 
fraction  may  be  reduced  to  a  decimal,  and  the  approximate  root  found. 

19.  What  is  the  square  root  of  |  =  '75  ?  Ans.  '866  -f-. 

20.  What  is  the  square  root  of  ££.  Ans.  '912  -f  . 

PRACTICAL.   EXERCISES    IX    THE    EXTRACTIOX    OF 
THE    SQUARE    ROOT. 

IF  210.      1.   A  general  has  4096  men  ;  how  many  must  he  place 
in  rank  and  file  to  form  them  into  a  square?  Ans.  64. 

2.  If  a  square  field  contains  2025  square  rods,  how  many  rods  does 
it  measure  on  each  side  ?  Ans.  45  rods. 

3.  How  many  trees  in  each  row  of  a  square  orchard  containing 
5625  trees?  Ans.  75. 

4.  There  is  a  circle  whose  area,  or  superficial  contents,  is  5184 
feet  ;  what  will  be  the  length  of  the  side  of  a  square  of  equal  area  ? 

^75184  =  72  feet,  Ans. 

5.  A  has  two  fields,  one  containing  40  acres,  and  the  other  con 
taining  50  acres,  for  which  B  offers  him  a  square  field  containing  the 


H210.  EVOLUTION.  267 

same  number  of  acres  as  both  of  these ;  how  many  rods  must  each 
side  of  this  field  measure  ?  Ans.  120  rods. 

6.  If  a  certain  square  field  measure  20  rods  on  each  side,  how 
much  will  the  side  of  a  square  field  measure,  containing  4  times  as 
much  ?  /\/20  X  20><T  =  40  rods,  Ans. 

7.  If  the  side  of  a  square  be  5  feet,  what  will  be  the  side  of  one  4 

times  as  large?  9  times  as  large1?  16  times  as  large?  

25  times  as  large?  36  times  as  large? 

Answers,  10  ft.  ;  15  ft.  ;  20  ft. ;  25  ft.,  and  30  ft. 

8.  It  is  required  to  lay  out  288  rods  of  land  in  the  form  of  a  paral 
lelogram,  which  shall  be  twice  as  many  rods  in  length  as  it  is  in 
width. 

NOTE  1.  —  If  the  field  be  divided  in  the  middle,  it  will  form  two  equal 
squares. 

Ans.  24  rods  long,  and  12  rods  wide. 

9.  I  would  set  out,  at  equal  distances,  784  apple  trees,  so  that  my 
orchard  may  be  4  times  as  long  as  it  is  broad ;  how  many  rows  of 
trees  must  I  have,  and  how  many  trees  in  each  row  ? 

Ans.  14  rows,  and  56  trees  in  each  row. 

10.  There  is  an  oblong  piece  of  land,  containing  192  square  rods, 
of  which  the  width  is  |  as  much  as  the  length ;  required  its  dimen 
sions.  %  Ans.  16  by  12. 

11.  There  is  a  circle,  whose  diameter  is  4  inches;  what  is  the 
diameter  of  a  circle  9  times  as  large  ? 

NOTE  2. —  A  square  4  inches  on  one  side,  contains  16  square  inches  ;  one 
twice  as  long,  or  8  inches  on  each  side,  contains  64  square  inches,  4  times  16  ; 
one  3  times  as  long,  or  12  inches  on  each  side,  contains  144  =9  times  16 
square  inches.  It  may  also  he  shown  by  geometry,  that  if  the  diameter  of  a 
circle  be  doubled,  its  contents  will  be  increased  4  times  ;  if  the  diameter  he 
trebled,  the  contents  will  be  increased  9  times.  That  is,  the  contents  of 
squares  are  in  proportion  to  the  squares  of  their  sides,  and  the  contents  of  cir 
cles  are  in  proportion  to  the  squares  of  their  diameters.  Hence,  to  perform 
the  above  example,  square  the  diameter,  multiply  the  square  by  9,  and  extract 
the  square  root  of  the  product. 

Ans.  12  inches. 

12.  There   are   two   circular   ponds   in   a   gentleman's    pleasure 
ground  ;  the  diameter  of  the  less  is  100  feet,  and  the  greater  is  3 
times  as  large  ;  what  is  its  diameter  ?  Ans.  173'2  -\-  feet. 

13.  If  the  diameter  of  a  circle  be  12  inches,  what  is  the  diameter 
of  one  i  as  large?  Ans.  6  inches. 

14.  A  carpenter  has  a  large  wooden  square;  one  part  of  it  is  4  feet 
long,  and  the  other  part  3  feet  long  ;  what  is  the  length  of  a  pole, 
which  will  just  reach  from  one  end  to  the  other  ? 

NOTE  3.  —  A  figure  of  3  sides  is  called  a  triangle,  and  if 
one  of  the  corners  be  a  square  corner,  or  right  angle,  like 
the  angle  at  B  in  the  annexed  figure,  it  is  called  a  right 
angled  triangle.  It  is  proved  by  a  geometrical  demonstra 
tion  that  the  square  contents  of  a  square  formed  on  the 
longest  side,  A  C,  are  equal  to  the  square  contents  of  the 
two  squares,  one  formed  on  each  of  the  other  two  sides,  A 
c  B,  and  C  B.  Thus,  Fig.  2,  a  square  formed  on  A  B,  the 
shortest  side,  will  co-  tain  9  square  feet,  the  square  on  C  B 


EVOLUTION. 


IT  210. 


Fig.  2. 


will  contain  16  square  feet,  9  -4-1(5  =  25 
square  feet,  in  bcth  squares.  The  square 
on  A  C  contains  25  small  squares  of  the 
same  size  as  the  squares  on  the  other  two 
sides  are  divided  into,  or  25  square  feet, 
and  the  square  root  of  25  will  be  the  length 
of  the  longest  side,  or,  Ans.,  5  feet. 
^  Hence,  if  the  length  of  the  two  short 
sides  are  given,  square  each,  add  the 
squares  tog-ether,  and  extract  the  square 
Square  root  of  the  sum  ;  the  root  will  be  the 
Length  of  the  long  side. 

If  the  long  side,  and  one  of  the  short 
sides  are  given,  square  each,  subtract  the 
square  of  the  short  side  from  the  square  of 
the  long  side ;  the  square  root  of  the  re 
mainder  will  be  the  other  short  side. 

EXAMPLES. 

15.  If,  from  the  corner  of  a  square  room,  6  feet  be  measured  off 
one  way,  and  8  feet  the  other  way,  along  the  sides  of  the  room,  what 
will  be  the  length  of  a  pole  reaching  from  point  to  point  ? 

Ans.  10  feet. 

16.  A  wall  is  32  feet  high,  and  a  ditch  before  it  is  24  feet  wide ; 
what  is  the  length  of  a  ladder  that  will  reach  from  the  top  of  the  wall 
to  the  opposite  side  of  the  ditch  ?  Ans.  40  feet. 

17.  If  the  ladder  be  40  feet,  and  the  wall  32  feet,  what  is  the  width 
of  the  ditch?  Ans.  24  feet. 

18.  The  ladder  and  ditch  given,  required  the  wall. 

Ans.  32  feet. 

19.  The  distance  between  the  lower  ends  of  two  equal  rafters  is  32 
feet,  and  the  hight  of  the  ridge,  above  the  beam  on  which  they  stand, 
is  12  feet ;  required  the  length  of  each  rafter.  Ans.  20  feet. 

20.  There  is  a  building  30  feet  in  length  and  22  feet  in  width,  and 
the  eaves  project  beyond  the  wall  1  foot  on  every  side  ;  the  roof  ter 
minates  in  a  point  at  the  centre  of  the  building,  and  is  there  supported 
by  a  post,  the  top  of  which  is  10  feet  above  the  beams  on  which  the 
rafters  rest ;  what  is  the  distance  from  the  foot  of  the  post  to  the  cor 
ners  of  the  eaves  ?  and  what  is  the  length  of  a  rafter,  reaching  to  the 

middle  of  one  side?  a  rafter  reaching  to  the  middle  of  one  end? 

and  a  rafter  reaching  to  the  corners  of  the  eaves  ? 

Answers,  in  order,  20  ft.  ;  15'62-f-ft. ;  18*86+  ft.  ;  and  22'36  + 
feet. 

21.  There  is  a  field  800  rods  long  and  600  rods  wide ;  what  is  the 
distance  between  two  opposite  corners  ?  Ans.  1000  rods. 

22.  There  is  a  square  field  containing-  90  acres  ;  hciw  many  rods 

Questions.  —  IT  210.  How  does  it  affect  the  contents  01  a  square  to 
double  its  length?  —  to  treble  its  length?  How  does  it  affect  the  contents 
of  circles  to  double  or  treble  their  diameters  ?  How  will  you  find  the  diam 
eter  of  a  circle  nine  times  as  large  as  one  of  a  given  diameter?  What  is  a 
right  angled  triangle  ?  What  is  said  of  the  squares  on  its  sides?  How  shown 
DV  Fig.  2  ?  When  both  short  sides  are  given,  how  do  you  find  the  long  side  ? 
When  the  long  side,  and  one  short  side  are  given,  how  do  you  find  the  ether? 


EVOLUTION. 


269 


in  length  is  each  side  of  the  field  ?  and  how  many  rods  apart  are  the 
opposite  corners?  Answers,  120  rods,  and  169'7-j-  rods. 

23.   There  is  a  square  field  containing  10  acres  ;  what  distance  is 
the  centre  from  each  corner?  Ans.  28'28 -\~ rods. 


Extraction  of  the  Cube  Root. 

1.     1.    How  many  feet  in  length  is  each  side  of  a 
cubic  block,  containing  125  solid  feet  ? 

SOLUTION.  —  As  the  solid  contents  of  a  cubical  body  are  found,  when 
one  side  is  known,  by  involving  the  side  to  the  third  power,  or  cube, 
(T[  206,)  so  when  the  solid  contents  are  known,  we  find  the  length  of 
one  side  by  extracting  the  cube  root,  a  number,  which,  taken  as  a  fac 
tor  3  times,  will  produce  the  given  number,  (^[  207.)  The  cube  root  of 
125  we  find  by  inspection,  or  by  the  table,  ^[  206,  to  be  5.  Ans.  5  feet. 

2.  What  is  the  side  of  a  cubic  block,  containing  64  solid  feet?  — 
27  solid  feet  ?  216  solid  feet?  512  solid  feet? 

Answers,  4  ft.,  3  ft.,  6  ft.,  and  8  ft. 

3.  Supposing  a  man  has  13824  feet  of  timber,  in  separate 
blocks  of  1  cubic  foot  each  ;  he  wishes  to  pile  them  up  in  a 
cubic  pile ;  what  will  be  the  length  of  each  side  of  such  a 
pile? 

SOLUTION.  —  It  is  evident  that,  as  in 
the  former  examples,  we  must  find 
the  length  of  one  side  of  a  cubical 
pile  which  13824  such  blocks  will 
make  by  extracting  the  cube  root  of 
13824.  But  this  number  is  so  large, 
that  we  cannot  so  easily  find  the  rcot 
as  in  the  former  examples ;  —  we  WL! 
endeavor,  however,  to  do  it  by  a  sort 
of  trial ;  and, 

1st.  We  will  try  to  ascertain  the 
number  of  figures,  of  which  the  root 
will  consist.  This  we  may  do  by 
pointing  the  number  off  into  periods 
of  three  figures  each.  For  the  cube 
of  any  figure  will  contain  3  times  as 
many,  or  1  or  2  less  than  3  times  as 
many  figures  as  the  number  itself. 
The  cube  of  2  contains  1  figure ;  the 
cube  of  5  contains  2  figures ;  the  cube 
af  9  contains  3  figures ;  the  cube  of  10 
contains  4  figures,  and  so  on. 

Pointing  off,  we  see  that  the  root 
will  consist  of  two  figures,  a  ten  and 
8000  feet,  contents.  a  unit.  Let  us,  then,  seek  for  the  first 

23* 


OPERATION. 

13824(2 

8 

5824 


270 


EVOLUTION. 


1211. 


figure,  or  tens  of  the  root,  which  must  be  extracted  from  the  left  hand 
period,  13,  (thousands.)  The  greatest  cube  in  13  (thousands)  we  find 
by  inspection,  or  by  the  table  of  powers,  to  be  8,  (thousands,)  the  root  of 
which  is  2,  (tens ;)  therefore,  we  place  2  (tens)  in  the  root.  As  the  root 
is  one  side  of  a  cube,  let  us  form  a  cube,  (Fi^.  I.,)  each  side  of  which 
shall  be  regarded  20  feet,  expressed  by  the  root  now  obtained.  The 
-,ontents  of  this  cube  are  20  X  20  X  20  =  8000  solid  feet,  which  are  now 
disposed  of,  and  which,  consequently,  are  to  be  deducted  from  the  whole 
number  of  feet,  13824.  8000  taken  from  13824  leave  5624  feet.  This 
deduction  is  most  readily  performed  by  subtracting  the  cubic  number,  8, 
or  the  cube  of  2,  (the  figure  of  the  root  already  found,)  from  the  period 
13,  (thousands,)  and  bringing  down  the  next  period  by  the  side  of  the 
remainder,  making  5824,  as  before. 

2d.  The  cubic  pile  A  D  is  now  to  be  enlarged  by  the  addition  of  5824 
solid  feet,  and,  in  order  to  preserve  the  cubic  form  of  the  pile,  the  addi 
tion  must  be  made  on  one  half  of  its  sides,  that  is,  on  3  sides,  a,  b,  and 
c.  Now  as  each  side  is  20  feet  square,  its  square  contents  are  400 
square  feet,  and  the  square  contents  of  the  3  sides  are  1200  square  feet. 
Hence,  an  addition  of  1  foot  thick  would  require  1200  solid  feet,  and 

dividing  5824  solid  feet  by  1200 
solid  feet,  the  contents  of  the  ad 
dition  1  foot  thick,  and  we  get  the 
thickness  of  the  addition.  It  will  be 
seen  that  the  quotient  figure  must 
not  always  be  as  large  as  it  can 
be.  There  might  be  enough,  for 
instance,  to  make  the  three  addi 
tions  now  under  consideration  5 
feet  thick,  when  there  would  not 
then  be  enough  remaining  to 
complete  the  additions. 

The  divisor,  1200,  is  contained 
in  the  dividend  4  times ;  conse- 


OPERATION- CONTINUED. 

13824(24500?. 

8 

Divisor,  1200)5824  Dividend. 

4800 

960 

64 


5824 
0000 


FIG.  II. 


quently,  4  feet  is  the  thickness  of 
the  addition  made  to  each  of  the 
three  sides,  a,  b,  c,  and  4  X  1209 
=  4800,  is  the  solid  feet  contained 
in  these  additions  ;  but  there  are 
still  1024  feet  left,  and  if  we  look 
at  Fig.  II.,  we  shall  perceive  that 
this  addition  to  the  3  sides  does 
not  complete  the  cube ;  for  there 
are  deficiencies  in  the  3  corners, 
n,  n,  n.  Now  the  length  of  each  of 
these  deficiencies  is  the  same  as 
the  length  of  each  side,  that  is,  2 
(tens)  =  20,  and  their  width  and 
thickness  are  each  equal  to  the  last 
quotient  figure,  (4 ;)  their  con 
tents,  therefore,  or  the  number  of 
feet  required  toy/7/  these  deficien 
cies,  will  be  found  by  multiply  ing 
the  square  of  the  last  quotient  fig 
ure.  C42J  =  16,  by  20 ;  16  X  20  = 


IT  212. 


EVOLUTION. 


271 


FIG.  111. 

20 


24/ee*. 


320  solid  feet,  required  for  one  de 
ficiency,  and  multiplying  320  by  3, 
320  X  3  =  969  so!icl  feet»  required 
for  the  3  deficiencies,  w,  n,  n. 

Looking  at  Fig.  III.,  we  perceive 
there  is  still  a  deficiency  in  the  cor 
ner  where  the  last  blocks  meet. 
This  deficiency  is  a  cube,  each  side 
of  which  is  equal  to  the  last  quo 
tient  figure,  4.  The  cube  of  4, 
therefore,  (4  X  4  X  4  =  64.)  will  be 
the  solid  contents  of  this  corner, 
which  in  Fig.  IV.  is  seen  filled. 

Now,  the  sum  of  these  several 
additions,  viz.,  4800  -j-  9(50  -f-  6.£ 
=±5824,  will  make  the  subtrahend, 
which,  subtracted  from  the  divi 
dend,  leaves  no  remainder,  and  the 
work  is  done. 

Fig.  IV.  shows  the  pile  which 
13824  solid  blocks  of  one  foot  each 
would  make,  when  laid  together, 
and  the  root,  24,  shows  the  length 
of  one  side  of  the  pile.  The  cor 
rectness  of  the  work  m?/  be  ascer 
tained  by  cubing  tlie  side  now 
found,  243,  thus,  24  X  24  X  24  = 
13824,  the  given  number ;  or  it 
may  be  proved  by  adding  together 
the  contents  of  all  the  several  parts, 
thus, 


Feet. 

8000  =  contents  of  Fig.  I. 

4800  =  addition  to  the  sides,  a,  b,  and  c,  Fig.  I. 

960  =  addition  to  fill  the  deficiencies  n,  n,  n,  Fig.  II. 
64  =  addition  to  fill  the  corner,  e,  e,  e,  Fig.  IV. 


13824  =  contents  of  the  whole  pile,  Fig.  IV.,  24  feet  on  each  side. 

IT  312.     From  the  foregoing  example  and  illustration  we 
ierive  the  following 


Questions. —  IT  211.  How  is  the  length  of  one  side  of  a  cube  found, 
when  the  contents  are  known  ?  Why,  Ex.  3,  is  the  number  pointed  off  as  it 
is?  How  many  figures  in  the  cube  of  any  number?  Illustrate  by  cubing 
*ome  numbers.  What  is  2,  the  first  figure  of  the  root  ?  Of  what  is  it  the 
root?  For  what  is  the  subtraction?  What  is  to  be  done  with  the  remain 
der?  On  how  many  sides  is  it  to  be  added,  and  why?  What  is  the  divisor, 
1200?  What  is  the  object  in  dividing  ?  The  quotient  expresses  what  ?  Why 
should  it  not  be  made  as  large  as  it  can  be  ?  What  additions  are  next  made, 
and  what  are  the  contents  of  each?  How  are  the  contents  found  ?  What  de 
ficiency  yei  remains,  and  how  large  ?  Of  what  parts  of  the  last  figure  does 
the  subtrahend  consist?  Describe  Fig.  I. ;  —  Fig.  II.  ;  —  Fig.  III.  ;  -  Fig. 
IV.  How  is  the  work  proved  ? 


272  EVOLUTION.  11212. 

RULE 

FOR   EXTRACTING   THE    CUBE    ROOT. 

I.  Place  a  point  over  the  unit  figure,  and  over  every  third 
figure  at  the  left  of  the  place  of  units,  thereby  separating  the 
given  number  into  as  many  periods  as  there  will  be  figures  in 
the  root. 

II.  Find  the  greatest  complete  cube  number  in  the  left 
hand  period,  and  place  its  cube  root  in  the  quotient. 

III.  Subtract  the  cube  thus  found  from  the  period  taken, 
and  bring  down  to  the  remainder  the  next  period  for  a  divi 
dend. 

IV.  Calling  the  quotient,  or  root  figure  now  obtained,  so 
many  tens,  multiply  its  square  by  3,  and  use  the  product  for 
a  divisor. 

V.  Seek  how  many  times  the  divisor  is  contained  in  the 
dividend,  and  diminishing  the  quotient,  if  necessary,  so  that 
the  whole  subtrahend,  when  found,  may  not  be  greater  than 
the  dividend,  place  the  result  in  the  root ;  then  multiply  the 
divisor  b^  this  root  figure,  and  write  the  product  under  the 
dividend. 

VI.  Multiply  the  square  of  this  root  figure  by  the  former 
figure  or  figures  of  the  root,  regarded  as  so  many  tens,  and 
the  resulting  product  by  3,  add  the  product  thus  obtained, 
together  with  the  cube  of  the  last  quotient,  to  the  former 
product  for  a  subtrahend. 

VII.  Subtract  the  subtrahend  from  the  dividend,  and  to  the 
remainder  bring  down  the  next  period  for  a  new  dividend, 
with  which  proceed  as  before,  till  the  work  is  finished. 

NOTE  1 .  —  If  it  happens  that  the  divisor  is  not  contained  in  the  dividend,  a 
cipher  must  be  put  in  the  root,  and  the  next  period  brought  down  for  a  divi 
dend. 

NOTE  2.  —  The  same  rule  must  be  observed  for  continuing  the  operation, 
nnd  pointing  off  for  decimals,  as  in  extracting  the  square  root. 

EXAMPLES    FOR    PRACTICE. 

4.    What  is  the  cube  root  of  1860867  ? 


Questions  —  IS  212,    What  is  the  general  rule?  —  note  1  ?  —  note  2? 
—  note  3  ? 


H  213.  EVOLUTION.  273 


OPERATION. 


1860867(123  Ans. 
I 


102  X  3  =  300 )  860  first  Dividend. 

600 
22X  10X3  =  120 

23   =     8 

728  first  Subtrahend. 


1202  X  3  =  43200)  132867  second  Dividend. 

129600 

32X  120x3=     3240 
33=         27 


132867  second  Subtrahend. 

000000 

5.  What  is  the  cube  root  of  373248  ?  Ans.  72. 

6.  What  is  the  cube  root  of  21024576?  Ans.  276. 

7.  What  is  the  cube  root  of  84'604519?  Ans..  4'39. 

8.  What  is  the  cube  root  of  '000343  ?  Ans.  '07. 

9.  What  is  the  cube  root  of  2?  Ans.  1'25-]-. 

10.  What  is  the  cube  root  of  -^  ?  Ans.  f  . 

NOTE  3.  —  The  cube  root  of  a  fraction  is  the  cube  root  of  the  numerate* 
divided  by  the  cube  root  of  the  denominator.     (IT  209.) 

11.  What  is  the  cube  root  of  £f  £  ?  Ans.  £  . 

12.  What  is  the  cube  root  of  y3^  ?  A™,  fa 

13.  What  is  the  cube  root  of  -^  ?  Ans.  '125  -f  . 

14.  What  is  the  cube  root  of  ----  ?  Ans.     . 


PRACTICAL,   EXERCISES    IN    EXTRACTING    THE    CUBE 
ROOT. 

If  SI  3.      1.    What  is  the  side  of  a  cubical  mound,  equal  to  one 
288  feet,  long,  216  feet  broad,  and  48  feet  high  ?         Ans.  144  feet. 

2.  There  is  a  cubic  box,  one  side  of  which  is  2  feet  ;  how  many 
solid  feet  does  it  contain  "?  Ans.  8  feet. 

3.  How  many  cubic  feet  in  one  8  times  as  large?  and  what  would 
be  the  length  of  one  sid  3  ? 

Ans.  64  solid  feet,  and  one  side  is  4  feet. 


274  EVOLUTION.  IT  214. 

4.  There  is  a  cubical  box,  one  side  of  which  is  5  feet ,  what  would 

be  the  side  of  one  containing  27  times  as  much  ? 64  times  as 

much?  125  times  as  much?  Ans.  15,  20,  and  25  feet. 

5.  There  is  a  cubical  box,  measuring  1  foot  on  each  side ;  what  is 
the  side  of  a  box  8  times  as  large?  27  times?  64  times? 

Ans.  2,  3,  and  4  feet. 

NOTE.  —  It  appears  from  the  above  examples  that  the  sides  of  cubes  are  as 
the  cube  roots  of  their  solid  contents,  and  their  solid  contents  as  the  cubes  ot 
their  sides.  It  is  also  true  that  if  a  globe  or  ball  have  a  certain  contents,  the 
contents  of  one  whose  diameter  is  double  are  8  times  as  great,  or  having  a 
treble  diameter  are  27  times  as  great,  and  so  on  ;  that  is,  the  contents  are  pro 
portional  to  the  cubes  of  their  diameters.  The  same  proportion  is  true  01  the 
similar  sides,  or  of  the  diameters  of  all  solid  figures  of  similar  forms. 

6.  If  a  ball,  weighing  4  pounds,  be  3  inches  in  diameter,  what  wiL 
be  the  diameter  of  a  ball  of  the  same  metal,  weighing  32  pounds  ? 
4  :  32  : :  33  :  63.  Ans.  6  inches. 

7.  If  a  ball  6  inches  in  diameter  weigh  32  pounds,  what  will  be  the 
weight  of  a  ball  3  inches  in  diameter?  Ans.  4  Ibs. 

8.  If  a  globe  of  silver,  1  inch  in  diameter,  be  worth  $6,  what  is  the 
value  of  a  globe  1  foot  in  diameter?  Ans.  $10368. 

9-  There  are  two  globes ;  one  of  them  is  1  foot  in  diameter,  and 
the  other  40  feet  in  diameter  ;  how  many  of  the  smaller  globes  would 
it  take  to  make  1  of  the  larger  ?  -4ns.  64000. 

10.  If  the  diameter  of  the  sun  is  112  times  as  much  as  the  diame 
ter  of  the  earth,  how  many  globes  like  the  earth  would  it  take  to 
make  one  as  large  as  the  sun?  Ans.  1404928. 

11.  If  the  planet  Saturn  is  1000  times  as  large  as  the  earth,  and 
the  earth  is  7900  miles  in  diameter,  what  is  the  diameter  of  Saturn  ? 

Ans.  79000  miles. 

12.  There  are  two  planets  of  equal  density ;  the  diameter  of  the 
less  is  to  that  of  the  larger  as  2  to  9  ;  what  is  the  ratio  of  their  solidi 
ties  ?  Ans.  Tf  -ff  ;  or,  as  8  to  729. 


IT  214.    Review  of  Involution  and  Evolution. 

Questions. — What  is  involution?  What  "are  powers?  How  are 
the  different  powers  represented?  How  is  a  number  involved?  What 
is  evolution  ?  What  is  a  root  ?  How  do  you  find  the  square  rcot,  01 
the  cube  root  of  a  number  ?  What  is  a  rational,  and  what  a  surd  num 
ber  ?  How  is  the  square  root  indicated  ?  —  the  cube  root  ?  Give  briefly 
the  solution  of  the  example  in  the  extraction  of  the  square  root ;  —  rule. 
How  are  decimals  pointed  off?  How  is  the  operation  continued,  when 
there  is  a  remainder?  Why  cannot  the  precise  root  be  ascertained? 
How  is  the  square  root  of  a  vulgar  fraction  found  ?  What  is  said  of  the 
relation  between  the  sides  and  contents  of  squares  ?  —  the  diameters  and 

Questions.  —  ^T  213.  What  proportion  exists  betweon  the  sides  of 
cubes,  and  their  solid  contents?  Illustrate.  What  betweer  the  diameters  of 
globes  and  their  contents  ?  If  you  increase  the  diameter  of  a  ball  5  times, 
now  much  are  its  contents  increased? 


IF  215.  ARITHMETICAL  PROGRESSION.  273 

contents  of  circles?  —  the  squares  on  the  sides  of  a  right-angled  trian 
gle  ?  Repeat  briefly  the  solution  of  the  example  in  cube  root ;  —  the 
rule.  What  is  said  of  the  relation  of  the  sides  of  cubes  to  the  contents  ? 
—  of  the  diameters  of  globes  to  their  contents  ? 

EXERCISES. 

1  What  is  the  difference  of  the  contents  of  6  fields,  each  20  rods 
square,  and  1  field  50  rods  square?  Ans.  100  square  rods. 

2.  What  is  the  difference  between  56  cubical  stacks  of  hay,  each 
10  feet  on  a  side,  and  1  stack  40  feet  on  a  side  ? 

Ans.  8000  solid  feet. 

3.  How  many  times  larger  is  a  circular  pond,  1  mile  in  diameter, 
than  one  that  is  40  rods  in  diameter?  Ans.  64  times. 

4.  What  is  one  side  of  a  cubical  pile  of  wood  which  contains  4 
cords?  Ans.  8  feet. 

5.  What  is  one  side  of  a  cubical  pile  of  bricks  which  will  lay  up 
the  walls  of  a  house  36  feet  high  and  16  inches  thick  for  the  first  12 
feet,  12  inches  the  next  12,  and  8  inches  the  upper  12,  the  house  being 
60  feet  long  and  34  wide  on  the  outside,  no  allowances  being  made 
for  windows,  doors,  &c.  ?     What  are  the  solid  contents  ? 

Ans.  to  the  last,  6613  cu.  ft.,  576  cu.  in. 

NOTE.  —  The  principal  object  in  evolution  is  to  find  one  side  of  a  square  or 
of  a  cube,  when  the  contents  are  known,  or  to  extract  the  square  and  cube 
roots.  There  are  methods  of  demonstrating  these  operations  different  from 
those  here  given,  which  are  preferable  in  some  respects,  but  they  are  deficient 
iu  one  important  particular — intelligibleness  to  those  for  whom  they  are  de 
signed.  In  a  "  higher  arithmetic"  they  might  be  appropriate. 

Other  roots  may  he  extracted  arithmetically,  but  the  methods  of  demonstrat 
ing  the  operations,  even  where  any  are  given,  are  difficult  of  comprehension. 
The  fourth  root,  however,  may  be  found  by  taking  the  square  root  of  the  square 
root,  the  sixth  root  by  taking"  the  square  foot  of  the  cube  root,  and  so  of  many 
other  roots.  Any  root  is  easily  taken  by  what  are  called  logarithms,  used  in 
the  more  advanced  departments  of  mathematics. 


ARITHMETICAL    PROGRESSION. 

IT  215.  1.  A  teamster  starts  with  5  barrels  of  flour ;  he 
passes  by  4  mills,  at  each  of  which  he  takes  on  3  barrels ; 
how  many  barrels  has  he  then  ? 

SOLUTION.  —  He  has  8  barrels  after  the  first  addition,  11  after  the 
second,  14  after  the  third,  and  17  after  the  fourth.  Ans.  17  barrels. 

2.  A  peddler  having  17  hats,  sold  3  at  each  of  4  stores , 
how  many  had  he  left? 

SOLUTION.  —  He  had  14  after  the  first  sale,  11  after  the  second.  8  after 
the  third,  and  5  after  the  fourth.  Ans.  5  hats. 

A  series  of  numbers  increasing  by  a  constant  addition,  or 


276  ARITHMETICAL   PROGRESSION.  11216. 

decreasing  by  a  constant  subtraction  of  the  sam«    number,  is 
called  an  Arithmetical  Progression. 

The  first  of  the  above  examples  is  called  an  ascending,  the 
second  a  descending  series. 

NOTE  1 .  —  The  numbers  which  form  the  series  are  called  the  terms  of  the 
series.  The  first  aud  last  terms  arc  the  extremes*  and  the  other  terms  are 
called  the  means. 

There  are  five  things  in  an  arithmetical  progression,  any  three  of  which 
being  given,  the  other  two  may  be  found :  — 

1st.  The  first  term. 

2d.  The  last  term. 

3d.  The  number  of  terms. 

4th.  The  common  difference. 

5th.  The  sum  of  all' the  terms. 

NOTE  2. —  The  common  difference  is  the  number  added  o:  subtracted  at  one 
time. 

5T  2 IB.  One  of  the  extremes,  the  common  difference,  and 
the  number  of  terms  being  given,  to  find  the  other  extreme. 

1.  A  man  bought  100  yards  of  cloth,  giving  4  cents  for  the 
first  yard,  7  cents  lor  the  second,  10  cents  for  the  third,  and 
so  on,  with  a  common  difference  of  3  cents;  what  was  -the 
cost  of  the  last  yard  ? 

SOLUTION.  —  We  add  3  to  4  cents,  (4  -4-  3  =  7,)  10  get  the  price  of  the 
second  yard,  3  to  7  to  get  the  price  of  the  third  yard,  and  so  on,  thus 
making  99  additions  to  4,  of  3  cents  each  ;  or,  we  may  take  3,  99  times, 
(the  multiplication  being  a  short  way  of  performing  the  99  additions,) 
and  add  the  product  to  4.  for  the  price"  of  the  last  yard,  3  X  99  5  or>  since 
either  factor  may  be  the  multiplier,  99  X  3  =  297,  and  44-297  =  301 
cents,  the  price  of  the  last  yard.  Ans.  301  cents. 

NOTE  1.  — The  prices,  4,  7,  10,  13  cents,  &c.,  are  an  ascending  series, 
which  has  as  many  terms  as  there  are  yards,  namely,  100  ;  3  is  the  common 
difference,  and  4  the  first  term,  to  which  99  times  3  must  be  added  to  find  the 
price  of  the  last  yard,  or  the  last  term.  It  is  added  1  time  less  than  the  num 
ber  of  terms,  since  4  is  the  price  of  the  first  yard  without  any  addition. 

Hence,  To  find  the  last  term  of  an  ascending  series  when 
the  first  term,  common  difference,  and  number  of  terms  are 
given, 

RULE. 

Multiply  the  common  difference  by  the  number  of  terms 
less  one,  to  get  the  sum  of  the  additions,  and  add  this  sum  to 
the  first  term ;  the  amount  will  be  the  last  term. 

Questions.  — IT  215.  How  is  Ex.  1  explained?  —  Ex.  2?  What  are 
the  extremes  ?  —  the  means  ?  —  the  terms  ?  How  many,  and  what  things 
are  there,  of  which,  if  three  are  given,  the  others  may  be  found  ?  What  is  the 
common  difference  ? 


11217.  ARITHMETICAL   PROGRESSION.  277 

NOTE  2.  —  If  the  same  things  are  given  of  a  descending  series,  we  must 
evidently  take  the  sum  of  the  subtractions  from  the  first  term  to  find  the  last. 
In  the  same  manner  we  may  find  the  first  term  of  an  ascending  series  when 
the  last  term  and  the  other  things  named  are  given  ;  but  having  thesr;  things 
given  of  a  descending  series,  we  find  the  first  term  by  the  rule  above  for  find 
ing  the  last  term  of  an  ascending  series. 

EXAMPLES    FOR    PRACTICE. 

2.  There  are  23  pieces  of  land,  the  first  containing  95  acres,  the 
second  91,  the  third  87,  and  so  on,  decreasing  by  a  common  difference 
of  4  ;  what  is  the  number  of  acres  in  the  last  piece?  Ans.  7. 

3.  The  first  term  of  a  series  is  6,  the  common  difference  is  3,  and 
the  number  of  terms  is  57 ;  what  is  the  last  term]  Ans.  174. 

4.  The  last  term  of  a  series  is  117,  the  common  difference  is  8, 
and  the  number  of  terms  is  15  ;  what  is  the  first  term?         Ans.  5. 

5.  The  last  term  is  6,  the  number  of  terms  21,  and  the  common 
difference  10 ;  what  is  the  first  term?  Ans.  206. 


Simple  Interest  by  Progression. 

1.    A  man  puts  out  $10,  at  6  per  cent.,  simple 
interest ;  to  what  does  it  amount  in  20  years  ? 

SOLUTION.  —  The  first  sum  is  $10,  the  amount  at  the  end  of  the  first 
year  is  $10*60,  at  the  end  of  the  second  year  Sil'20,  increasing  each 
year  by  the  constant  addition  of  $'60.  Hence,  simple  interest  is  a  case 
of  arithmetical  progression,  the  principal  being  the  first  term,  the  inter 
est -for  one  year  being  the  common  difference,  the  number  of  terms  one 
more  than  the  number  of  years,  rince  there  is  one  term,  the  principal, 
at  the  commencement  of  the  first  year,  and  one  term,  the  amount  for  a 
year,  at  its  close,  and  the  last  term,  \vhich  we  wish  to  find,  is  the  amount 
for  the  number  of  years.  To  find  the  last  term,  or  this  amount,  multi 
ply  the  interest  for  1  year  by  the  number  of  years,  (one  less  than  the  num 
ber  of  terms,)  and  add  the  product  to  t/te  first  term.  Ans.  $22. 

2.  Two  lads,  at  14  years  of  aj?e,  commence  labor  for  themselves ; 
the  one  lays  up  nothing,  but.  the  oilier,  by  prudence,  lays  up  $300  by 
the  time  he  is  20  years  old,  which  he  puts  out  at  7  per  cent,  simple 
interest ;  afterwards,  each  earns  his  living,  and  no  more  ;  at  the  age 
of  70  the  one  is  worth  nothing,  and  comes  upon  public  charity  ;  what 
is  the  other  worth  at  that  age?  Ans.  $1350. 

Questions.  —  IT  216.  What  things  are  given,  and  what  are  required,  in 
IT  21€- .'  How  many  cases  may  there  be,  and  what  are  they?  What  are 
given  L.  Ex.  1?  Ho\v  much  is  the  first  term  increased  to  make  the  last? 
Why  ai  .  oulj  09  times  3  added  ?  Give  the  rule.  To  what  case  does  it  apply? 
What  is  done  i  i  each  case,  when  other  things  are  given? 

IT  217.  H  w  does  it  appear  that  simple  interest  is  a  case  of  progression? 
What  things,  according  to  IT  216,  are  given,  and  what,  is  required?  Why  is 
there  one  more  term  than  the  number  of  years  ?  HOM  is  simple  interest  per 
formed  by  progression  ? 

24 


278  ARITHMETICAL  PROGRESSION.        IF  218, 219. 

3.  What  will  a  watch,  purchased  at  21  for  $25,  cost  an  individual 
by  the  time  he  is  75,  reckoning  nothing  for  repairs  but  simple  interest 

at  6  per  cent,  on  the  purchase  money?  at  8  per  cent.  1 

Ans.  to  the  last,  $133. 

5T  218.  The  extremes  and  the  number  of  terms  given,  to 
faid  the  common  difference. 

1.  The  prices  of  100  yards  are  in  arithmetical  progression, 
the  first  being  4,  the  last  being  301  cents ;  what  is  the  com 
mon  increase  of  price  on  each  succeeding  yard? 

SOLUTION.  —  As  the  first  yard  costs  4  cents,  297  cents  have  been 
added  to  4  for  the  price  of  the  last  yard,  at  99  times,  and  dividing  the 
number  added  at  99  times  by  99,  wo  get  the  number  added  at  1  time. 
Hence, 

RULE. 

Divide  the  whole  number  added  or  subtracted,  by  the  num 
ber  of  additions  or  subtractions,  that  is,  the  difference  of  the 
extremes  by  the  number  of  terms  less  1,  and  the  quotient  i? 
the  number  added  or  subtracted  at  one  time,  or  the  common 
difference, 

EXAMPLES  FOR  PRACTICE. 

2.  If  the  extremes  be  5   and  605,  and  the  number  of  terms  151, 
what  is  the  common  difference?  Ans.  4. 

3.  A  man  had  8  sons,  whose  ages  differed  alike  ;  the  youngest  wa? 
10  years  old,  and  the  eldest  45  ;  what  was  the  common  difference  of 
their  ages?  Ans.  5  years. 

NOTE.  —  If  the  extremes  and  common  difference  are  given,  we  may  find  the 
number  of  terms  by  dividing  the  difference  of  the  extremes  by  the  common 
difference,  and  adding  1  to  the  quotient. 

4.  The  extremes  are  5  and  1205,  and  the  common  difference  8; 
what  is  the  number  of  terms?  Ans.  151. 

^[  219.  The  extremes  and  the  number  of  terms  being 
given,  to  find  the  sum  of  all  the  terms. 

1.  What  is  the  amount  of  the  ascending  series,  3,  5,  7,  9 
11,  13,  15,  17,  19? 

SOLUTION.  —  The  sum  may  be  found  by  adding  together  the  terms  , 
but  in  an  extended  series  this  process  would  be  tedious.  We  will  there 
fore  seek  for  a  shorter  method ;  and  first,  will  write  down  the  terms  of 

Questions.  —  IT  218.  What  are  given  and  what  required,  IT  218  ?  Ex 
plain  how  the  common  difference  is  found,  Ex.  1.  Give  the  rule.  How  is 
the  number  of  terms  found,  wh  an  the  extremes  and  the  common  difference  are 
given  7 


IF  220.  ARITHMETICAL    PROGRESSION.  279 

the  series  in  order,  and  beginning  with  the  last,  write  the  terms  of  the 
same  series  under  these,  placing  the  last  term  under  the  first,  the  next 
to  the  last  under  the  second,  the  third  from  the  last  under  the  third,  and 
so  on,  thus  :  — 

3   5   7   9  11  13  15  17  19 
19  17  15  13  11   9   7   5   3 

22  22  22  22  22  22  22  22  22 

Adding  together  each  pair,  we  see  that  the  sums  are  alike,  and  the 
amount  of  the  whole  is  as  many  times  22,  the  first  sum,  as  there  are 
terms  in  either  series,  which  is  9.  22  X  9  =  198,  the  number  in  both 
series,  and  198  -j-  2  =  99  must  be  the  sum  of  the  first  series,  which  we 
wish  to  find.  But  22  is  the  sum  of  the  extremes  of  the  series  ;  hence, 
when  the  extremes  and  the  number  of  terms  are  given  to  find  the  sum  0}  the 
terms, 

RULE. 

Multiply  the  sum  of  the  extremes  by  the  number  of  terms, 
and  half  the  product  will  be  the  sum  of  the  terms. 

EXAMPLES  FOR  PRACTICE. 

2.  If  the  extremes  be  5  and  605,  and  the  number  of  terms  151, 
what  is  the  sum  of  the  series?  Ans.  46055. 

3.  What  is  the  sum  of  the  first  100  numbers,  in  their  natural 
order,  that  is,  1,  2,  3,  4,  &c.  1  Ans.  5050. 

4.  How  many  times  does  a  common  clock  strike  in  12  hours'? 

Ans.  78. 


Annuities  by  Arithmetical  Progression. 

IT  22O.  An  annuity,  (from  the  Latin  word  annus,  mean 
ing  a  year,)  is  a  uniform  sum,  due  at  the  end  of  every  year. 
When  payment  is  not  made  at  the  end  of  the  year,  the  annu 
ity  is  said  to  be  in  arrears,  and  the  sums  of  the  annuities 
should  draw  interest  just  as  any  other  debts  not  paid  when 
due.  When  on  simple  interest,  the  several  years'  annuities, 
with  the  interest  on  each,  form  an  arithmetical  progression, 
and  the  calculation  to  ascertain  the  whole  sum  due,  is  — 
finding  the  sum  of  an  arithmetical  series;  thus  :  — 

Questions.  —  IT  219.  What  are  given,  and  what  is  required,  IF  219? 
How  might  the  s  im  be  found?  What  difficulty  in  this  method?  What  is 
ihe  process  by  wh  ch  a  shorter  method  is  found  7  What  fact  do  we  discover 
from  tne  additions,  Ex.  1  ?  What  does  the  product  express,  and  why  is  it 
divided  by  2  ?  What  is  the  quotient  ?  Why  is  22  tho  sum  of  the  extremes  1 
Give  the  rule. 


280  ARITHMETICAL  PROGRESSION.  IT  221. 

1.  A  man,  whose  salary  is  $100  a  year,  does  not  receive 
anything-  till  the  end  of  8  years  ;  what  was  then  his  due,  sim 
ple  interest  on  the  sums  in  arrears  at  6  per  cent.  ? 

SOLUTION.  —  The  first  year's  salary  not  being  paid  till  7  years  after 
it  is  due,  since  it  was  due  at  the  end  if  the  first  year,  is  on  interest  7 
years.  The  interest  of  $100,  7  years,  is  $42,  and  $  100  -j-  42  =  8142, 
which  he  should  receive  on  account  of  his  first  year's  salary.  His  sec 
ond  year's  salary,  on  interest  6  years,  will  amount  to  $136,  his  third 
year's  salary  will  amount  to  $130,  and  so  on,  decreasing  uniformly 
by  $6,  the  interest  of  $100  for  a  year,  till  the  last  year,  when  the  salary, 
being  paid  at  the  end  of  the  year  for  which  it  has  accrued,  will  not  be 
on  interest,  but  will  yield  him  $100.  The  sums,  $142,  $136,  $130, 
$124,  $118,  $112,  $106,  $100,  form  a  descending  arithmetical  progres 
sion,  and  to  find  the  sum  due,  multiply  the  sum  of  the  extremes  by  the 
number  of  terms,  and  take  half  the  product,  thus  :  — 

S142-J- SI 00  =  $242 ;  and  8242  X  8  =  $1936.  which  -f-  2  =  $968,  Ans. 
EXAMPLES    FOR    PRACTICE. 

2.  A  soldier  of  the  revolution  did  not  establish  his  claim  to  a  pen 
sion  of  $96  a  year  till  10  years  after  it  should  have  begun  ;  what  was 
then  his  due,  simple  interest  on  the  sums  in  arrears  at  6  per  cent.  ? 

Ans.  $1219'20. 

3.  A  man  uses  tobacco  at  an  expense  of  $5  a  year  from  the  age  of 
18  till  the  age  of  79,  when  he  dies,  leaving  to  his  heirs  $300 ;  what 
might  he  have  left  them  if  he  had  dispensed  with  the  worse  than  use 
less  article,  and  loaned  the  money,  which  it  cost  him,  at  the  end  of 
each  year,  for  7  per  cent.,  simple  interest?  Ans.  $1245'50. 

4.  A  and  B  have   the  same  income,  but  the  expenses  of  A  are 
$210  a  year,  and  those  of  B  are  $250 ;  at  the  end  of  40  years  B  is 
worth  $1500  ;  what  is  A  worth,  having  loaned  what  he  saved  more 
than  B  at  7  per  cent,  simple  interest  at  the  end  of  each  year  ? 

Ans.  $5284. 

5.  Refer  to  ^f  164,  Ex.  8  :  what  would  the  merchant  gain  if  he 
continued  in  trade  31  years  to  borrow  money  instead  of  purchasing  on 
credit,  loaning  the  money  saved  at  the  end  of  each  year  at  7  per  cent, 
simple  interest!  Ans.  $20151'70  nearly. 

EXERCISES. 

IT  5*^1.  1.  If  a  triangular  piece  of  land,  30  rods  in  length,  be 
20  rods  wide  at  one  end,  and  come  to  a  point  at  the  other,  what  num 
ber  of  square  rods  does  it  contain  1  Ans.  300. 

Questions.  —  'I  220.  What  is  an  annuity  ?  Why  so  called  ?  When 
are  annuities  in  arrears'?  Why  should  they  then  draw  interest?  When  will 
they  form  an  arithmetical  progression?  What  is  the  calculation  to  find  the 
whole  stun  due  /  Why,  Ex.  1,  was  the  first  year's  salary  on  interest  7  years  ? 
Show  how  long  each  year's  salary  was  on  interest.  Why  was  not  the  last 
year's  salary  on  interest  ?  Why  do  the  sums  form  a  descending  series?  How 
nay  the  sum  be  found?  Why"  is  not  the  number  of  terms  ona  more  than  the 
number  of  years,  as  in  IT  217  ? 


11222.  GEOMETRICAL    PROGRESSION.  281 

2.  A  debt  is  to  be  discharged  at  11  several  payments,  in  arithmeti 
cal  series,  the  first  to  be  $5,  and  the  last  $75 ;  what  is  the  whole 
debt  ?  the  common  difference  between  the  several  payments  1 

Arts.  Whole  debt,  $440 ;  common  difference,  $7 

3.  What  is  the  sum  of  the  series  1,  3,  5,  7,  9,  &c.,  to  1001'? 

:    .  Ans.  251001. 

NOTE.  —  The  number  of  terms  must  first  be  found. 

4.  A  man  bought  100  yards  of  cloth  in  arithmetical  series;  he 
gave  4  cents  for  thejirst  yard,  and  301  cents  for  the  last  yard  ;  what 
was  the  amount  of  the  whole?  Ans.  $152'50. 

5.  What  annuity,  in  20  years,  at  6  per  cent,  simple  interest,  will 
amount  to  $1570  ?  Ans.  $50. 

6.  What  is  the  sum  of  the  arithmetical  series,  2,  2£,  3,  3£,  4,  4£, 
&c.,  to  the  50th  term  inclusive'?  Ans.  712£. 

7.  What  is  the  sum  of  the  decreasing  series,  30,  29|,  29 J,  29, 
28f,  &c.,  down  to  0?  Ans.  1365. 

8.  A  laboring  female  was  able  to  put  $30,  at  the  end  of  each  year, 
in  the  savings  bank,  at  5  per  cent.,  simple  interest,  from  the  age  of 
18  till  77,  when  she  died  ;  how  much  had  she  become  worth  ? 

Ans.  $4336*50. 


GEOMETRICAL   PROGRESSION. 

IT  222.  1.  A  man,  having  5  acres  of  land,  doubled  the 
quantity  at  the  end  of  each  year  for  4  years;  how  many  acres 
had  he  then  ? 

SOLUTION.  —  Having  5  acres  at  first,  he  had  2  times  5,  or  10,  at  the 
end  of  the  first  year,  2  times  10,  or  20,  at  the  end  of  the  second  year,  2 
times  20,  or  40,  at  the  end  of  the  third  year,  and  2  times  40,  or  80,  at  the 
end  of  the  fourth  year.  Ans.  80  acres. 

2.  A  lady,  having  $80,  traded  at  4  stores,  expending  one 
half  at  the  first,  half  of  what  she  had  left  at  the  second,  and 
so  on,  expending  half  the  money  in  her  possession  at  each 
store  till  the  last :  what  had  she  left  ? 

SOLUTION.  —  She  leaves  the  first  store,  to  which  she  came  with  $80 
-:-  2  =  $40,  the  second  with  $40  -f-  2  =$20,  the  third  with  $20  —  2  = 
$10,  the  fourth  with  $10  -5-  2  =$5.  Ans.  $5. 

Any  series  of  numbers  like  5,  10,  20,  40,  80,  increasing  by 
the  same  multiplier,  or  like  80,  40,  20,  10,  5,  decreasing  by 
the  same  divisor,  is  called  a  geometrical  progression. 

The  multiplier  or  divisor  is  called  the  ratio.    The  first  and 
last  terms  are  called  the  extremes. 
24* 


282  GEOMETRICAL  PROGRESSION.  U  223. 

The  first  is  called  an  increasing,  the  second  is  called  a  de 
creasing  geometrical  series. 

NOTE. — As  in  arithmetical,  so  also  in  geometrical  progression,  there  are 
five  things,  any  three  of  which  being  given,  the  other  two  may  be  found :  — 
1st.  Theirs*  term. 
2d.  The  last  term. 
3d.  The  number  of  terras. 
4th.  The  ratio. 
5th.  The  sum  of  all  the  terms. 

IT  923.  The  first  term,  number  of  terms,  and  ratio  of  an 
increasing  geometrical  series  being  given,  to  find  the  last  term. 

1.  A  man  agreed  to  pay  for  13  valuable  houses,  worth 
$5000  each,  what  the  last  would  amount  to,  reckoning  7  cents 
for  the  first,  4  times  7  cents  for  the  second,  and  so  on,  in 
creasing  the  price  4  times  on  each  to  the  last ;  did  he  gain  or 
lose  by  the  bargain,  and  how  much  ? 

SOLUTION. — We  multiply  7  cents,  the  sum  reckoned  for  the  first 
house,  by  4,  to  get  the  sum  reckoned  for  the  second  house,  and  this  by 
4  to  get  the  sum  reckoned  for  the  third  house,  and  so  on,  multiplying  12 
times  by  4.  But  to  multiply  twice  by  4  is  the  same  as  to  multiply  once 
by  16.  which  is  the  second  power  of  4,  or  4  times  4.  Thus,  7  X  4  =  28, 
and  28  X  4  =  112.  So,  also,  7  X  16=  112.  Hence,  multiplying  7  by 
the  twelfth  power  of  4  is  the  same  as  multiplying  by  4  12  times.  And 
412,  that  is,  the  twelfth  power  of  4,  is  16777216;  and  7  X  16777216, 
("using,  if  we  choose,  the  larger  factor  for  the  multiplicand,  ^[  21,)  pro 
duces  117440512  cents,  =  $117440542  ;  the  houses  were  worth  $5000  X 
13  =865000,  and  $1174405'12  —  $65000  =  $1109405<12,  loss,  Ans. 

Hence,  when  the  first  term,  number  of  terms,  and  ratio  of 
an  increasing  series  are  given,  to  find  the  last  term, 

RULE. 

Multiply  the  first  term  by  the  ratio  raised  to  a  power  one 
less  than  the  number  of  terms. 

NOTE  1 .  —  To  get  a  high  power  of  a  number,  it  is  convenient  to  write  down 
a  few  of  the  lower  powers,  and  multiply  them  together,  thus  :  powers  of  4. 

1st  power.    2d  power.    3d  power.    4th  power.    5th  power.    6th  power.     7th  power. 

4    16    64   256   1024  4096  16384,  &c. 

Now  the  7th  power,  multiplied  by  the  5th  power,  will  produce  the  12th  power, 
as  also  the  6th  by  the  4th,  and  the  product  by  the  2d  ;  the  5th  by  the  4th,  and 
the  product  by  the  3d  ;  the  7th  by  the  4th,  and  the  product  by  the  1st,  &c 
Multiplying  all  the  powers  now  written,  will  produce  the  28th  ;  all  but  the 
5th,  will  produce  the  23d  ;  all  but  the  2  1,  will  produce  the  2Gth,  &c. 

Questions.  —  IF  222.  What  is  a  geometrical  progression?  — an  in 
creasing  series?  —  a  decreasing  series?  — the  extremes?  —  the  ratio  1 
What  rive  things  are  there,  of  which,  if  three  are  given,  the  others  may  be 
found 


f  224.  GEOMETRICAL  PROGRESSION  283 

EXAMPLES    FOR    PRACTICE. 

2.  A  man  plants  4  kernels  of  corn,  which,  at  harvest,  produce  32 
kernels  :  these  he  plants  the  second  year ;  now,  supposing  the  annual 
increase  to  continue  8  fold,  what  would  be  the  produce  of  the  15th 
year,  allowing  1000  kernels  to  a  pint! 

NOTE  2.  —  The  4  kernels  planted  is  the  first  term,  and  the  32  kernels  lar- 
vested  the  second  term,  both  within  the  first  year. 

Ans.  2199023255<552  bushels. 

3.  Suppose  a  man  had  put  out  one  cent  at  compound  interest  in 
1620,  what  would   have  been  the  amount  in  1824,  allowing  it  to 
double  once  in  12  years?  217=  131072.     Ans.  $1310*72. 

NOTE  3.  —  When  the  ratio,  the  number  of  terms,  and  the  last  term  of  a  de 
creasing  series  is  given,  the  first  term  is  evidently  found  by  the  same  rule. 
But  when  the  same  things  are  given  of  a  decreasing  series,  as  those  of  the 
ascending  series  required  by  the  rule,  that  is,  the  first  term,  ratio,  and  number 
of  terms,  we  must  divide  the  first  term  by  the  ratio  raised  to  a  power  which  19 
one  less  than  the  number  of  terms.  In  the  same  way  we  may  find  the  first 
term  of  an  increasing  series,  when  the  ratio,  number  of  terms,  and  last  term 
are  given. 

4.  If  the  last  term  of  a  decreasing  series  be  5,  the  ratio  3,  and  the 
number  of  terms  7,  what  is  the  first  term?  Ans.  3645. 

5.  If  the  first  term  of  a  decreasing  series  be  10935,  the  ratio  3, 
and  the  number  of  terms  8,  what  is  the  last  term?  Ans.  5. 

6.  If  the  last  term  of  an  increasing  series  be  196608,  the  number 
of  terms  17,  and  the  ratio  2,  what  is  the  first  term?  Ans.  3. 

NOTE  4.  —  When  the  first  and  last  terms,  and  the  ratio  are  given,  to  find  the 
number  of  terms,  we  may  divide  the  greater  term  by  the  less,  the  quotient  by 
the  ratio,  and  so  on,  continually  dividing  by  the  ratio  till  nothing  remains  • 
the  number  of  divisions  will  be  equal  to  the  number  of  terms. 

7.  The  first  term  is  7,  the  ratio  10,  and  the  last  term  700000000  ; 
what  is  the  number  of  terms  ?  Ans.  9. 


Compound  Interest  by  Progression. 

IT  294.  1.  To  what  will  $40  amount  in  4  years,  com 
pound  interest  at  6  per  cent.  ? 

Questions.  —  IT  223.  What  are  given,  and  what  is  required,  in  If  223  ? 
How  is  the  last  term  found,  as  first  described?  What  may  be  done  instead 
of  this?  Why?  How  may  a  high  power  be  obtained?  How  the  20tn 
power?  —the  16th  power?  —the  27  .h  power?  —the  llth  power?  What 
w  the  rule  ?  Give  the  substance  of  no  ,e  3 ;  —  of  note  4. 


284 


GEOMETRICAL  PROGRESSION 


H224. 


OPERATION. 

$40',  prin.,  or  1st  term 
1'06 


42<40,  2d  term. 
1'06 

25440 
42*40 


44'9440,  3d  term. 
1*06 

2696640 
449440 


47'640640,  4th  term. 
106 


285843840 
47640640 

50-49907840,  5th  term. 


SOLUTION.  —  The  amount  of 
$40  for  one  year  is  once  $40  -f. 
T&rof  $40,  orl<06  (i$£)  of 
$40,  and  to  obtain  it,  we  mul 
tiply  $40  by  1<06.  This  pro 
duct,  multiplied  by  1'06,  gives 
the  amount  for  2  years.  Hence, 
compound  interest  is  a  case  of 
an  increasing  geometrical  pro 
gression,  of  which  there  are 
given  the  first  term,  or  princi 
pal,  the  ratio,  or  the  amount  of 
$1  for  1  year,  and  the  number 
of  terms,  which  is  one  more 
than  the  number  of  years,  there 
being  2  terms  the  first  year, 
the  principal  at  the  commence 
ment,  and  the  amount  with  one 
year's  interest  at  the  close  ;  and 
we  are  to  find  the  last  term  — 
the  amount  for  the  time  —  by 
the  rule  in  the  last  ^[.  The 
several  terms,  it  will  be  seen, 
increase  by  the  common  multi 
plier,  1<06. 

Ans.  $50<499-f. 

NOTE  1.  —  The  powers  of  the 
ratio  may  usually  be  found  in  the 
table,  11  161,  since  they  are  the 
same  with  the  amounts  of  $1  for 
the  number  of  years  which  indi 
cates  the  ower  of  the  ratio  that 


we  wish.  It  appears  also  that  the  only  difference  in  finding  the  amount  of  a 
sum  at  compound  interest  by  the  table,  and  by  progression,  is  the  order  in 
which  we  take  the  factors.  By  If  161,  we  multiply  the  amount  of  SI  for  the 
time  by  the  number  of  dollars  ;  by  progression,  we  multiply  the  number  of 
dollars  by  the  ratio  raised  to  a  power  denoted  by  the  number  of  years,  or  the 
amount  of  $1  for  the  time. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  amount  of  40  dollars,  for  11  years,  at  5  per  cent., 
compound  interest  ?  Ans.  $68'413-j— 

3  What  is  the  amount  of  $6,  for  4  years,  at  10  per  cent.,  com 
pound  interest  1  Ans.  $8'  784-^. 

4.  In  what  time  will  $1000  amount  to  $1191'016,  at  6  per  cent., 
compound  interest? 

NOTE  2.  —  The  case  is  evidently  one  of  finding  the  number  of  terms,  (one 
more  than  the  number  of  years,)  when  the  ratio  and  the  first  and  last  terms 
are  given,  IT  223,  note  4.  Ans.  3  years. 

Questions.  —  IT  224.  How  does  it  appear  from  the  illustration  that  com 
pound  interest  is  a  case  of  progression  ?  What  are  the  things  given  ?  —  to  find 
what  ?  What  is  the  ratio,  Ex.  1  ?  Why  ?  How  may  tlje  powers  ol  the  ratio  usu 
ally  be  found  ?  Why  ?  What  does  it  appear  are  gnen,  and  what  is  required, 
Ex.4? 


H  225,  226.         GEOMETRICAL  PROGRESSION. 


285 


Compound  Discount. 

IT  225.  1.  What  is  the  present  worth  of  $304*899,  due 
4  years  hence  without  interest,  money  being  worth  6  per 
cent,  compound  interest  ? 

SOLUTION.  —  We  find,  by  the  table,  f  161,  that  SI,  in  4  years,  amounts 
to  $  1 '26247,  and  as  many  times  as  this  amount  of  $1  is  contained  in 
the  given  sum,  so  many  dollars  it  will  be  worth ;  for  it  is  worth  a  sum, 
which,  put  at  compound  interest  4  years,  would  amount  to  it,  and  divid 
ing  the  amount  of  the  number  of  dollars  by  the  amount  of  one  dollar, 
—  we  have  the  number  of  dollars,  or,  Ans.  241*509  -\-. 

NOTE.  —  The  case  is  evidently  one  of  a  geometrical  progression,  in  which 
the  ratio,  (1[06,)  the  number  of  terms,  (5,)  and  the  greater  term  are  given,  to 
find  the  less,  as  in  IT  223,  note  3. 

TABLE, 

Showing  the  present  worth  of  $1,  or  £1,  from  1  year  to  40, 
allowing  compound  discount,  at  5  and  6  per  cent. 


Years. 

5  per  cent. 

6  per  cent. 

Years. 

5  per  cent. 

6  per  cent. 

1 

'952381 

'943396 

21 

'358942 

'294155 

2 

'907029 

'889996 

22 

'341850 

'277505 

3 

'863838 

'839619 

23 

'325571 

'261797 

4 

'822702 

'792094 

24 

'310068 

'246979 

5 

'783526 

'747258 

25 

'295303 

'232999 

6 

'746215 

'704961 

26 

'281241 

'219810 

7 

'710881 

'665057 

27 

'267848 

'207368 

8 

'676839 

'627412 

28 

'255094 

'195630 

9 

'644609 

'591898 

29 

'242946 

'184557 

10 

'613913 

'558395 

30 

'231377 

'174110 

11 

'584679 

'526788 

31 

'220359 

'164255 

12 

'556837 

'496969 

32 

'209S66 

'154957 

13 

'530321 

'468839 

33 

'199873 

'146186 

14 

'505068 

'442301 

34 

'190355 

'137912 

15 

'481017 

'417265 

35 

'181290 

'130105 

16 

'458112 

'393646 

36 

'172657 

'122741 

17 

'436297 

'371364 

37 

'164436 

'115793 

18 

'415521 

'350344 

38 

'156605 

'109239 

19 

'395734 

'330513 

39 

'149148 

'103056 

20 

'376889 

'311805 

40 

'142046 

'097222 

IF  22O.     The  extremes  and  the  ratio  given  to  find  the  sum 
of  the  series. 


Questions.  —  IT  225.    What  is  compound  discount  ?    How  is  the 
ent  vrorth  found  ?    Like  what  case  in  a  geometrical  progression  is  it  ? 


286  GEOMETRICAL  PROGRESSION.  11"  226 

I 

1.  A  man  bought  4  yards  of  cloth,  giving  2  cents  for  the 
first  yard,  6  for  the  second,  18  for  the  third,  and  54  for  the 
fourth ;  what  does  he  pay  for  all  ? 

SOLUTION. — We  may  add  together  the  prices  of  the  several  yards 
thus: 

2  _|-  6  -J-18  +  54  =  80. 

But  in  a  lengthy  series,  this  process  would  be  tedious ;  we  will  therefore 
seek  for  a  shorter  method.  Writing  down  the  terms  of  the  series,  we 
multiply  the  first  term  by  the  ratio,  and  place  the  product  over  the  sec 
ond,  to  which  it  will  be  equal,  since  the  second  term  is  the  product  of  the 
first  into  the  ratio.  Multiply,  also,  the  second  term,  placing  the  product 
over  its  equal,  the  third ;  multiply  the  third,  placing  the  product  over  the 
fourth  j  multiply  the  fourth,  and  place  the  product  at  the  right  of  the  last 
product,  thus : 

Second  series,       6     18     54     162 
First  series,     2     6     18    54 

000 
162 
2 


2)  160,  twice  the  first  series. 

80,  sum  of  the  first  series. 

The  second  series  is  three  times  the  first  series,  and  subtracting  the 
first  from  it,  there  will  remain  t\vice  the  first  series.  But  the  terms  bal 
ance  each  other,  except  the  first  term  of  the  first  series,  the  sum  of  which 
we  wish  to  find,  and  the  last  term  of  the  second,  which  is  3  times  the  last 
term  of  the  series  whose  sum  we  wish.  _  Subtracting  the  former  from 
the  latter,  we  have  left  160,  twice  the  sum  of  the  first,  which  dividing 
by  2,  the  quotient  is  80,  sum  of  the  series  required.  Hence, 

RULE. 

Multiply  the  larger  term  by  the  ratio,  and  subtract  the  less 
term  from  the  product,  divide  the  remainder  by  the  ratio  less 
1 ;  the  quotient  will  be  the  sum  of  the  series. 

EXAMPLES  FOR  PRACTICE. 

2.  If  the  extremes  be  4  and  131072,  and  the  ratio  8,  what  is  the 
sum  of  the  series  ?  Ans.  149790. 

3.  What  is  the  sum  of  the  decreasing  series,  3,  1,  },  £,  2\,  &c 
extended  to  infinity  ? 

Questions.  —  IT  226.  What  are  given,  to  find  w'\at,  IT  226  ?  How  might 
the  sum  be  found  ?  What  difficulty  in  this  ?  What  is  the  manner  of  proceed" 
ing  to  find  a  shorter  method  ?  Give  the  rule.  Explain  the  reason  of  the  rule 
What  is  an  infinite  series,  and  what  its  last  term? 


IT  227.  GEOMETRICAL    PROGRESSION.  287 


—  such  a  series  is  called  an  infinite  series,  the  last  term  of  which  is 
so  near  nothing  that  we  regard  it  0  ;  hence  when  the  extremes  are  3  and  0, 
and  the  ratio  3,  what  is  the  sum  of  the  series  1 

Ans.  4£. 

4.  What  is  the  value  of  the  infinite  series,  1  -\-  J,  -j-  yV  ~t~  ^V* 
&c.  ?  Ans.  1J. 

5.  What  is  the  value  of  the  infinite  series,  -£$  -f"  Ttrtf  +  uniT) 
+  Ttf?rnj>  &c.,  or>  what  is  the   same>  the  decimal   '11111,  &c.. 
continually  repeated?  Ans.  ^. 

6.  What  is  the  value  of  the  infinite  series,  yg^  -|-  TOoVtr»  &c-> 
decreasing  by  the  ratio  100,  or,  which  is  the  same,  the  repeating 
decimal  '020202,  &c.  ?  Ans.  -/§- 

IF  227.  Thejirst  term,  ratio,  and  number  of  terms  given 
to  find  the  sum  of  the  series. 

1.  A  lady  bought  6  yards  of  silk,  agreeing  to  pay  5  cents 
for  the  first  yard,  15  for  the  second,  and  so  on,  increasing  in 
a  three  fold  proportion  ;  what  did  the  whole  cost  ? 

SOLUTION.  —  We  may  find  the  prices  of  the  several  yards,  and  add 
them  together,  or,  having  found  the  last  term  by  If  223,  we  can  find  the 
sum  by  the  last  ^j.  But  our  object  is  to  find  a  still  more  expeditious 
method.  Let  us  find  the  several  terms  and  write  them  down  as  a  first 
series,  and  below  it  write  a  series  which  we  will  call  the  second,  having 
1  for  the  first  term,  and  the  same  number  of  terms,  thus  : 

First  series,       5     15     45     135    405     1215 

6th  power 

Third  series,  3       9       27       81       243    °729°' 

Second  series,     1       3       9       27       81       243 

729  —  1  =  728,  which  -j-  2  =  364,  and  364  X  5  =  1820  cents. 

Now  multiplying  the  second  series  by  the  ratio,  3,  and  writing  the  prod 
ucts  as  directed  in  the  last  If,  we  have  a  series  three  times  the  second 
The  last  term  of  the  third  series,  it  must  be  carefully  noticed,  is  the  6th 
power  of  3,  the  ratio,  the  power  denoted  by  the  number  of  terms.  Sub 
tracting  the  second  series  from  the  third,  which  is  done  by  taking  1  from 
the  last  term  of  the  third,  the  other  terms  balancing,  729  —  1  =  728,  we 
have  twice  the  second  series,  and  dividing  728  by  2,  728  -f-  2  =  364,  we 
have  once  the  second  series.  Now  the  first  series,  the  sum  of  which  is 
required,  is  5  times  the  second,  since,  as  the  first  term  is  5  times  greater, 
each  term  is  5  times  greater  than  the  corresponding  term  of  the  second 
series  ;  and  multiplying  364.  the  sum  of  the  second,  by  5,  we  have  the 
required  sum,  or  1820  cents  =  $18'20,  Ans. 

Hence,  the  first  term,  ratio,  and  number  of  terms  leing 
given,  to  find  the  sum  of  the  series, 

RULE. 

Raise  the  ratio  to  a  power  whose  index  is  equal  to  the 


288  GEOMETRICAL    PROGRESSION.  11228. 

number  of  terms,  from  which  subtract  1,  and  divide  the  re 
mainder  by  the  ratio  less  1  ;  the  quotient  is  the  sum  of  a 
series  with  1  for  the  first  term;  then  multiply  this  quotient  by 
the  first  term  of  any  required  series  ;  the  product  will  be  its 
amount. 

EXAMPLES  FOR  PRACTICE. 

2.  A  gentleman,  whose  daughter  was  married  on  a  new  year's 
day,  gave  her  a  dollar,  promising  to  triple  it  on  the  first  day  of  each 
month  in  the  year  ;  to  how  much  did  her  portion  amount? 

Applying  this  rule  to  the  example,  312  =  531441,  and 

1  =  265720.  Ans.  $265,720. 

3.  A  man  agrees  to  serve  a  farmer  40  years  without  any  other  re 
ward  than  1  kernel  of  corn  for  the  first  year,  10  for  the  second  year, 
and  so  on,  in  tenfold  ratio,  till  the  end  of  the  time  ;  what  will  be  the 
amount  of  his  wages,  allowing  10CO  kernels  to  a  pint,  and  supposing 
he  sells  his  corn  at  50  cents  per  bushel1? 

1040—  1  v     _J  1,111,111,111,111,111,111,111,111, 
10—1  X          |         111,111,111,111,111  kernels. 
Ans.  $8,  680,  555,  555,  555,  555,  555,  555,  555,  555,  555,  555'555-^W 

4.  A  gentleman,  dying,  left  his  estate  to  his  5  sons  ;  to  the  young 
est  $1000,  to  the  second  $1500,  and  ordered  that  each  son  should  ex 
ceed  the  younger  by  the  ratio  of  1<|  ;  what  was  the  amount  of  the 
estate  ? 

NOTE.  —  Before  finding  the  power  of  the  ratio  1£,  it  may  be  reduced  to  an 
improper  fraction  =  £,  or  to  a  decimal,  1'5. 

X  1000  =  $13187J;  or,  \f~*  X  1000  =  $13187'50, 


Answer. 


Annuities  at  Compound  Interest. 

IT  3S8.  1.  A  man  rented  a  dwelling-house  for  $100  a 
year,  but  did  not  receive  anything  till  the  end  of  4  years, 
when  the  whole  was  paid,  with  compound  interest  at  6  per 
cent.,  on  the  sums  not  paid  when  due';  what  did  he  receive? 

Questions.  —  IT  227.  What,  is  the  first  method  of  finding  the  sum,  when 
the  things  are  given,  named  in  IT  227?  —  the  second?  Describe  the  process 
for  finding  a  third  method.  What  terms  constitute  the  first  series  ?  —  the 
second  ?  How  is  the  third  found  ?  What  do  you  say  of  its  last  term  ?  How 
does  it  appear  to  be  so  ?  How  much  is  the  remainder,  after  subtracting  the 
second  from  the  third  series  ?  How,  then,  is  the  sum  of  the  second  series 
found  ?  How  the  sum  of  the  first,  and  why  ?  Give  the  rule.  Why  raise 
tlie  ratio,  &c.  ? 


IT  229  GEOMETRICAL  PROGRESSION.  289 

SOLUTION.  —  As  annuities  in  arrears  at  simple  interest  lorm  an  arith 
metical  series,  so  the  several  years'  rents  with  compound  interest  on 
those  in  arrears,  are  so  many  terms  of  a  geometrical  series.  The  last 
year's  rent  is  $100  only,  since  it  is  paid  when  due,  at  the  end  of  the 
year ;  the  third  year's  rent  is  on  interest  1  year,  and  is  found  by  mulli- 

?  lying  $100  by  1<06,  producing  $106',  and  this  product  multiplied  by 
'06,  will  give  the  second  year's  rent,  paid  2  years  after  it  is  due,  arid  so 
on.     The  first  term,  $100,  the  number  of  terms,  4,  and  the  ratio,  1<06, 
are  given  to  find  the  sum,  as  in  the  last  If,  and  we  may  apply  the  same 
rule,  thus :  — 

~'°f04~-  X  100  =  437<45.  Ans.  $437<45. 

NOTE.  —  The  powers  of  the  ratio,  see  IT  224,  may  be  found  in  the  table, 
IT  161. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  amount  of  an  annuity  of  $50,  it  being  in  arrears 
20  years,  allowing  5  per  cent,  compound  interest  ? 

Ans.  $1653*29. 

3.  If  the  annual  rent  of  a  house,  which  is  $150,  be  in  arrears  4 
years,  what  is  the  amount,  allowing  10  per  cent,  compound  interest  1 

Ans.  $696' 15. 

4.  To  how  much  would  a  salary  of  $500  per  annum  amount  in  14 
years,  the  money  being  improved  at  6  per  cent,  compound  interest? 

in  10  years?  in  20  years?  in  22  years?  in  24 

years?  Ans.  to  the  last,  $25407'75. 

5.  Two  men  commence  life  together  ;  the  one  pays  cash  down, 
$200  a  year  to  mechanics  and  merchants ;  the  second  gets  precisely 
the  same  value  of  articles,  but  on  credit,  and  proving  a  negligent  pay 
master,  is  charged  20  per  cent,  more  than  the  other ;  what  is  the  dif 
ference  in  40  years,  compound  interest  being  calculated  at  6  per  cent.  ? 

Ans.  $6 190*478 +. 

6.  A  family  removes  once  a  year  for  30  years,  at  an  expense  and 
loss  of  $100  each  time  ;  what  is  the  amount,  6  per  cent,  compound 
interest  being  calculated  ?  Ans.  $7905*818 -|-. 


Present  Worth  of  Annuities  at  Compound 
Interest. 

1T  229.  1.  A  man,  dying-,  left  to  his  nephew,  21  years 
old,  the  use  of  a  house,  which  would  rent  at  8300  a  year 
for  10  years,  after  which  it  was  to  come  in  the  possession 
of  his  own  children ;  the  young  man,  wishing  ready  money 
to  commence  business  in  a  small  shop,  rented  the  house  for  10 

Questions.  —  IT  228.  How  does  it  appear  that  an  annuity  is  an  example 
of  geometrical  progression  ?  Why  is  the  number  of  terms  only  equal  to  the 
number  of  years?  What  is  to  be  found,  and  by  what  rule? 

25 


290 


GEOMETRICAL  PROGRESSION. 


IF  230. 


years,  receiving  in  advance  such  a  sum  as  was  equivalent  to 
$300  paid  at  the  end  of  each  year,  reckoning  compound  dis 
count  at  6  per  cent. ;  what  did  he  receive  ? 

SOLUTION.  —  First,  we  find  what  he  would  receive  at  the  end  of  10 
years,  if  nothing  had  been  paid  before,  by  the  last  ^[.  Now  what  he 
should  receive  at  the  commencement  of  the  10  years,  is  a  sum,  which, 
on  compound  interest  at  the  rate  given,  would  amount  to  this  in  10 
years,  and  we  divide  it  by  the  amount  of  $1,  found  as  above,  for  the 
present  worth.  Ans.  $2208*024. 

EXAMPLES    FOR    PRACTICE. 

2.  What  is  the  present  worth  of  an  annual  pension  of  $100,  to 
continue  4  years,  allowing  G  per  cent,  compound  interest ' 

Ans.  $346<503-|-. 

3.  What  is  the  present  worth  of  an  annual  salary  of  $100,  to  con 
tinue  20  years,  allowing  5  per  cent.  1  Ans.  $1246*218  -[-. 

IT  23O.  The  operations  under  this  rule  being  somewhat 
teiious,  we  subjoin  a 

TABLE, 

Showing  the  present  worth  of  $1  or  £1  annuity,  at  5  and  6 
per  cent,  compound  interest,  for  any  number  of  years  from 
1  to  40. 


Years. 

5  per  cent. 

6  per  cent. 

Years. 

5  per  cent. 

6  per  cent. 

1 

0<95238 

0*94339 

21 

12*82115 

11'76407 

2 

1>85941 

1<S3339 

22 

13463 

12'04158 

3 

2'72325 

2*67301 

23 

13*48807 

12'30338 

4 

3*54595 

3<4(>51 

24 

13'79864 

12'55035 

5 

4*32948 

4*21236 

25 

14*09394 

12'78335 

6 

5107569 

4*91732 

26 

14*37518 

13-00316 

7 

#78637 

5*58238 

27 

U'64303 

13'21053 

8 

6*46321 

6'20979 

28 

14*89813 

13*40616 

9 

7'10782 

6'80169 

29 

1544107 

13*59072 

10 

7'72173 

7'36008 

30 

15*37245 

13*76483 

11 

8*30641 

7<SS6S7 

31 

15'59281 

13*92908 

12 

8*86325 

S'38384 

32 

15*80268 

14*08398 

13 

9-39357 

8*85268 

33 

16'00255 

14'22917 

14 

9*89864 

9*29493 

34 

164929 

14'36613 

15 

10*37986 

9*71225 

35 

16*37419 

14'49S24 

16 

10*83777 

1040589 

36 

16*54685 

14*62098 

17 

11*27407 

10*47726 

37 

16'71128 

14'73678 

18 

11'68958 

10*3276 

38 

16'86789 

14*84601 

19 

12*08532 

H'15811 

39 

17'01704 

14'94907 

20 

12'46221 

11*46992 

40 

1745908 

15^04629 

1T231.  GEOMETRICAL  PROGRESSION.  291 

NOTE  1.  — From  the  table  it  appears  that,  instead  of  $1  a  year  for  30  years, 
paid  at  the  end  of  each,  which  would  be  $30,  one  would  receive  at  the  com 
mencement,  S15'37245,  at  5  per  cent.,  or  $13'76483,  at  6  per  cent,  compound 
discount,  and  for  $50  a  year,  50  times  as  much.  Hence,  for  finding  the  pres 
ent  worth  at  compound  discount  by  the  table, 

RULE. 

Multiply  the  present  worth  of  $1  by  the  number  of  dollars. 

EXAMPLES    FOR    PRACTICE. 

1.  What  ready  money  will  purchase  an  annuity  of  $150,  to  con 
tinue  30  years,  at  5  per  cent,  compound  interest? 

Ans.  $2305'8675. 

2.  What  is  the  present  worth  of  a  yearly  pension  of  $40,  to  cor 

tinue  10  years,  at  6  per  cent,  compound  interest? at  5  per  cent.  ? 

to  continue  15  years  ?  20  years  1  25  years  1  34 

years?  Ans.  to  last,  $647'716. 

NOTE  2.  —  The  practised  arithmetician  will  have  no  difficulty  in  calculating 
the  present  worth  of  annuities  at  simple  interest,  from  principles  heretofore 
presented. 


Annuities  at  Compound  Interest  in  Reversion. 

If  231*  NOTE.  —  An  annuity  is  said  to  be  in  reversion  when  it  does 
not  commence  immediately. 

1.  In  Ex.  1,  IF  229,  supposing  the  uncle  had  reserved  the 
use  of  the  house  to  his  sister  for  2  years  after  the  young  man 
was  21,  and  given  it  to  him  for  10  years  after  this  time  should 
have  expired,  how  much  could  he  have  obtained  with  which 
to  commence  business  ? 

SOLUTION.  —  If  he  should  wait  till  he  is  23  years  old,  he  could  obtain 
$2208'024,  as  already  found,  and  he  can,  at  21,  obtain  a  sum  which,  at 
compound  interest,  would  amount,  in  two  years,  to  $2208*024,  or  the 
present  worth  of  this  sum  paid  two  years  before  due,  found  by  Tf  225  to 
be  Ans.  $1965<13-[-. 

Hence,  to  find  the  present  worth  of  an  annuity  in  reversion, 

RUL.E. 

Find  the  present  worth,  were  it  to  commence  now,  and  the 
present  worth  of  this  sum  for  the  time  in  reversion. 

Questions.  -  IT  229.    Giye  the  first  example.    What  sum  shduld  he 

receive  now '?    How  is  it  found? 

TT  23O.  What  appears  from  the  table?  How  is  the  present  worth  of  $50 
found?  Rule. 


292  GEOMETRICAL  PROGRESSION.  IF  232 

EXAMPLES    FOR   PRACTICE. 

2.  What  ready  money  will  purchase  the  reversion  of  a  lease  of  $60 
per  annum,  to  continue  6  years,  but  not  to  commence  till  the  end  of  3 
years,  allowing  6  per  cent,  compound  interest  to  the  purchaser  ? 

The   present   worth,   to   commence   immediately,   we   find  to  be 


$295*039,  and  -p  =  247'72.  Ans.  $247<72. 

3.  What  is  the  present  worth  of  $100  annuity,  to  be  continued  4 
years,  but  not  to  commence  till  2  years  hence,  allowing  0  per  cent. 
compound  interest?  Ans.  $308'392-|- 

4.  What  is  the  present  worth  of  a  lease  of  $100,  to  continue  20 
years,  but  no*  to  commence  till  the  end  of  4  years,  allowing  5  per 
cent.  ?  -  what,  if  it  be  6  years  in  reversion  ?  -  8  years  ?  - 
10  years?  -  14  years?  Ans.  to  last,  $629*426. 

5.  The  revolutionary  war  closed  in  1783  ;  one  of  the  soldiers  com 
menced  receiving,  in  1817,  a  pension  of  $96  a  year,  which  continued 
till  1840  ;  what  was  the  pension  worth  to  him  at  the  close  of  the  war, 
the  rate  being  6  per  cent,  compound  interest  ?       Ans.  $162'  89  -+-. 


Perpetual  Annuities, 

IT  239.  1.  A  farm  rents  for  $60  a  year,  at  6  per  cent. ; 
what  is  its  value  ? 

SOLUTION.  —  This  is  a  perpetual  annuity,  since  the  owner  is  supposed 
to  receive  $60  a  year  forever.  On  every  dollar  which  the  farm  is  worth 
he  receives  6  cents,  and  consequently  the  farm  is  worth  as  many  dollars 
as  the  number  of  times  6  cents  are  contained  in  $60.  $60-i-$<U6 
=  $1000,  Ans. 

Hence,  to  find  the  worth  of  a  perpetual  annuity, 

RULE. 

Divide  the  annuity  by  the  rate  per  cent. ;  the  quotient  will 
De  the  perpetual  annuity. 

2.  A  city  lot  is  rented  999  years,  at  $800  a  year ;  what  is  it  worth, 
the  rate  being  7  per  cent.  ? 

NOTE  1.  — This  is  the  same  as  a  perpetual  annuity.          Ans.  $1142S'57-}-. 

3.  What  is  the  worth  of  $100  annuity,  to  continue  forever,  allow 
ing  to  the  purchaser  4  per  cent.  ?  allowing  5  per  cent.  ?  8 

per  cent.  ?  10  per  cent.  ?  15  per  cent. '  20  per  cent.  • 

Ans.  to  last,  $500. 

4.  A  farm  is  left  me  which  will  rent  for  $60  a  year,  but  is 

Questions.  —  IT  231.  What  do  you  understand  by  annuities  in.  rever 
«on  ?  How  is  the  worth  of  an  annuity  in  reversion  found  ? 


IT  233.  PERMUTATION.  293 

not  to  come  into  my  possession  till  the  end  of  2  years  ;  what 
is  it  worth  to  me,  the  rate  being  6  per  cent,  compound  inter* 
est? 

SOLUTION.  —  The  farm  will  be  worth  $1000  to  me  2  years  hence,  and 
it  is  now  worth  a  sum  which,  put  at  compound  interest  2  years,  will 


amount  to  $1000.  =  $889<996,  Ans. 

5.  What  is  the  present  worth  of  a  perpetual  annuity  of  $100,  to 
commence  6  years  hence,  allowing  the  purchaser  5  per  cent,  com 
pound  interest?  -  what,  if  8  years  in  reversion?  -  10  years? 
-  4  years?  -  15  years?  -  30  years  ? 

Ans.  to  last,  $462*755. 

NOTES.  —  The  foregoing  examples,  in  compound  interest,  have  been  con 
fined  to  yearly  payments  ;  if  the  payments  are  half  yearly,  we  may  take  half 
the  principal  or  annuity,  half  the  rate  per  cent.,  and  ticice  the  number  of  years, 
and  work  as  before,  and  so  for  any  other  part  of  a  year. 


PERMUTATION. 

IT  233.  Permutation  is  the  method  of  finding;  how  many 
different  ways  the  order  of  any  number  of  things  may  be  va 
ried  or  changed. 

1.  Four  gentlemen  agreed  to  dine  together  so  long  as  they 
could  sit,  every  day,  in  a  different  order  or  position;  how 
many  days  did  they  dine  together  ? 

SOLUTION.  —  Had  there  been  but  two  of  them,  a  and  b,  they  could  sit 
only  in  2  times  1  (1  X2  =  2)  different  positions,  thus,  a  b,  and  b  a. 
Had  there  been  three,  a,  b,  and  c,  they  could  sit  in  1  X  2  X  3  =  6  differ 
ent  positions  ;  for,  beginning  the  order  with  a,  there  will  be  2  positions, 
viz.,  a  b  c,  and  a  c  b ;  next,  beginning  with  b,  there  will  be  2  positions, 
b  a  c,  and  b  c  a  ;  lastly,  beginning  with  c,  we  have  c  a  b,  and  c  b  a,  thai 
is,  in  all,  1X2X3  =  6  different  positions.  In  the  same  manner,  if 
there  be  four,  the  different  positions  will  be  1  X  2  X  3  X  4  =  24. 

Ans.  24  days. 

Hence,  to  find  the  number  of  different  changes  or  permuta 
tions,  of  which  any  number  of  differe?it  things  is  capable, — 
Multiply  continually  together  all  the  terms  of  the  natural 

Questions.  —  IT  232.  What  do  you  understand  by  a  perpetual  annuity  ? 
How  is  its  value  found  ?  Rule.  When  it  does  not  begin  immediately,  how 
is  its  worth  calculated  ?  What  do  you  say  of  other  than  yearly  uayments  ? 

IT  233.  What  is  permutation  ?  Illustra'.e  by  the  first  example.  What  it 
the  rule  ? 

25* 


294  MISCELLANEOUS  EXAMPLES.  IF  234 

series  of  numbers,  from  1  up  to  the  given  number,  and  the 
last  product  will  be  the  answer. 

2.  How  many  variations  may  there  be  in  the  position  of  the  nine 
digits?  Ans.  362880. 

3.  A  man  bought  25  cows,  agreeing  to  pay  for  them  1  cent  foi 
every  different  order  in  which  they  could  all  be  placed ;  how  much 
<ti  the  cows  cost  him  ?  Ans.  $1551 1210043330985984000000. 

i.    Christ  Church,  in  Boston,  has  8  bells ;  how  many  changes  may 
be  rung  upon  them  ?  Ans.  40320. 


MISCELLANEOUS    EXAMPLES. 


1T234.     1.  7-f  4  —  2-f  3-f  40X5  =  howmany?    Ans.  230. 

NOTE.  —  A  line  drawn  over  several  numbers,  signifies  that  the  whole  are  to 
be  taken  as  one  number. 

2.  The  sum  of  two  numbers  is  990,  and  their  difference  is  90 ;  what 
are  the  numbers  ? 

3.  There  are  4  sizes  of  chests,  holding  respectively  48,  76,  87  and  90 
Ibs. ;  what  is  the  least  number  of  pounds  of  tea  that  will  exactly  fill  some 
number  of  chests  of  either  of  the  4  sizes  ?  Ans.  396720  Ibs. 

4.  How  many  bushels  of  wheat,  at  $1*50  per  bushel,  must  be  given 
for  15  yards  of  cloth  worth  2s.  3d.  sterling  per  yard '? 

Am.  5^(yQ  bushels. 

5.  If  oats,  worth  $'30  per  bushel,  are  sold  for  $'35  on  account,  for 
what  ought  cloth  to  be  sold  on  account,  worth  $3<75  per  yard  cash? 

Ans.  $4<37£. 

5.  Bought  a  book,  marked  $4<50,  at  33  J  per  cent,  discount  for  cash  ; 
what  did  I  pay  ?  Ans.  $3<00. 

7.  Bought  120  gallons  of  molasses  for  $42  j  how  must  I  sell  it  per 
gallon  to  gain  15  per  cent.  ?  Ans.  $<40£. 

8.  What  sum,  at  6  per  cent,  interest,  will  amount  to  $150  in  2  years 
and  6  months  ?  Ans.  $130<434  +. 

9.  What  is  the  present  worth  of  $1000,  payable  in  4  years  and  2 
months,  discounting  at  the  rate  of  6  per  cent.  ?  Ans.  $800. 

10.  Bought  cloth  at  $3<50  per  yard,  and  sold  it  for  $4<25  per  yardj 
what  did  I  gain  per  cent.  ?  Ans.  21^  per  cent. 

11.  If  20  men  can  build  a  bridge  in  60  days,  how  many  would  be  re 
quired  to  build  it  in  50  days  ?  Ans.  24  men. 

12.  How  much  Silesia,  1£  yards  wide,  will  line  12  yards  of  plaid,  g 
yd.  wide  ?  Ans.  5  yards. 

13.  A  cistern,  holding  400  gallons,  is  supplied  by  a  pipe  at  the  rate 
of  7  gallons  in  5  minutes,  but  2  gallons  leak  out  in  6  minutes ;  in  what 
time  will  it  be  filled  ?  Ans.  6  hours  15  minutes. 

14.  A  ship  has  a  leak  which  would  cause  it  to  sink  in  10  hours,  but 
it  could  be  cleared  by  a  pump  in  15  hours  j  in  what  time  would  it  sink  ? 

Ans.  30  hours. 

15.  How  long  must  I  keep  $300,  to  balance  the  use  of  $500,  which  1 
lent  a  friend  4  months  ?  Ans.  6|  month* 


U  234.  MISCELLANEOUS  EXAMPLES.  295 

6.  If  800  men  have  provisions  for  2  months,  how  many  must  leave 
that  the  remainder  may  subsists  months  on  the  same?          Ans.  480. 

17.  Bought  45  barrels  of  beef,  at  $3'50  per  barrel,  except  16  barrels, 
for  4  of  which  I  pay  no  more  than  for  3  of  the  others  ;  what  do  the  whole 
cost?  Ans.  $143<50. 

18.  A  hare,  running  36  rods  a  minute,  has  57  rods  the  start  of  a  dog ; 
how  far  must  the  dog  run  to  overtake  him,  running  40  rods  per  minute? 

Ans.  570  rods. 

19.  The  hour  and  minute  hands  of  a  watch  are  together  at  12  o'clock ; 
when  are  they  next  together?  Ans.  1  h.  5  m.  27T3Ts.  P.  M. 

20.  Three  men  start  together  to  travel  the  same  way  around  an  island 
20  miles  in  circumference,  at  the  rate  of  2,  4,  and  6  miles  per  hour ;  in 
what  time  will  they  be  together  again  ?  Ans.  10  hours 

21.  Two  boats,  propelled  by  steam  engines  8  miles  an  hour,  start  at 
the  same  time,  the  one  up,  the  other  down  a  river,  from  places  300  miles 
apart ;  at  what  distance  from  the  place  where  each  started  will  they 
meet,  if  the  one  is  retarded,  and  the  other  accelerated  2  miles  an  hour 
by  the  current  ? 

Ans.  112£  miles  from  the  lower,  187£  from  the  upper  place. 

22.  The  third  part  of  an  army  were  killed,  the  fourth  part  taken  pris 
oners,  and  1000  fled  ;  how  many  in  the  army?  Ans.  2400. 

23.  A  farmer  has  his  sheep  in  5  fields  :  £  in  the  first,  £  in  the  second, 
£  in  the  third,  -fa  m  the  fourth,  450  in  the  fifth ;  how  many  sheep  has 
he?  Ans.  1200. 

24.  If  a  pole  be  &  in  the  mud,  f  in  the  water,  and  6  feet  out  of  the 
water,  what  is  its  length?  Ans.  90  feet. 

25.  If  T^-  of  a  school  study  grammar,  f  geography,  -fa  arithmetic, 
^j-  learn  to  write,  and  9  read,  what  number  in  the  .school  ?     Ans.  80. 

26.  A  man  being  asked  how  many  geese  he  had,  replied,  if  I  had  £ 
as  many  as  I  now  have,  and  2J  geese  more,  added  to  my  present  num 
ber,  I  should  have  100  ;  how  many  had  he?  Ans.  65. 

27.  In  a  fruit  orchard,  £  the  trees  bear  apples,  |  pears,  |-  plums,  100 
peaches  and  cherries  ;  how  many  in  all  ?  Ans.  1200. 

28.  The  difference  between  £  and  f  of  a  number  is  6  j  required  the 
number.  Ans.  80. 

29.  What  number  is  that,  to  which,  if  ^  and  £  of  itself  be  added,  the 
sum  will  be  84  ?  Ans.  48. 

30.  B's  age  is  1£  times  the  age  of  A,  C's  age  2^  times  the  age  of 
both,  and  the  sum  of  their  ages  is  93  years ;  required  the  age  of  each. 

Ans.  A  12  years,  B  18  years,  C  63  years. 

31.  If  a  farmer  had  as  many  more  sheep  as  he  now  has,  |,  &,  |,  and 
&  as  many,  he  would  have  435  ;  how  many  has  he?  Ans.  120. 

32.  Required  the  number,  which,  being  increased  by  |  and  f  of  itself, 
aud  by  22,  will  be  three  times  as  great  as  it  now  is.  Ans.  30. 

33.  A  and  B  commence  trade  with  equal  sums ;  A  gained  a  sum 
equal  to  £  of  his  stock,  B  lost  $200.  when  A's  money  was  twice  B's  ; 
what  stock  had  each  ?  Ans.  $500. 

34.  A  man  was  hired  50  days,  receiving  $'75  for  every  day  he 
worked,  and  forfeiting  $<25  for  every  day  he  was  idle;  he  received 
$27<50  ;  how  many  days  did  he  work  ?     .  Ans.  40. 

35.  A  and  B  have  the  sams  income ;  A  saves  £  of  his ;  B,  spending 


296  MISCELLANEOUS  EXAMPLES.  1T  234. 

$30  a  year  more  than  A,  is  $40  in  debt  at  the  end  of  8-  years ;  what  did 
B  spend  each  year  ?  Ans.  $205. 

36.  A  man  left  to  A  £  his  property,  wanting  $20,  to  B  4,  to  C  the 
rest,  which  was  $10  less  than  A's  share ;  what   did  each  receive? 

Ans.  A  received  $80,  B  $50,  C  $70. 

37.  The  head  of  a  fish  is  4  feet  long,  the  tail  us  long  as  the  head  and 
£  the  length  of  the  body,  the  body  as  long  as  the  head  and  tail ;  what  is 
the  length  of  the  fish  ?  Ans.  32  feet. 

38.  A  can  build  a  wall  in  4  days,  B  in  3  days ;  in  what  time  can 
both  together  build  it  ?  Ans.  1$  days. 

39.  A  and  B  can  build  a  wall  in  4  days,  B  and  C  in  6  days,  A  and  C 
in  5  days ;  required  the  time  if  they  work  together.      Ans.  3-^y  days. 

40.  A  and  B  can  build  a  wall  in  5  days ;  A  can  build  it  in  7  days; 
in  how  many  days  can  B  build  it?  Ans.  17^  days. 

41.  A  man  left  his  two  sons,  one  14,  the  other  18  years  old,  $1000, 
so  divided,  that  their  shares,  being  put  at  6  per  cent,  interest,  should  be 
equal  when  each  should  be  21  years  old ;  what  was  the  share  of  each? 

Ans.  8546'153-J-;  $453<S46-f-. 

42.  What  is  paid  for  the  rent  of  a  house  5  years,  at  $60  a  year,  in 
arrears  for  the  whole  time  at  6  per  cent,  simple  interest? 

Ans.  $336. 

43.  If  3  dozen  pairs  of  gloves  be  equal  in  value  to  40  yards  of  calico, 
and  100  yards  of  calico  to  90  yards  of  satinet,  worth  $'50  a  yard,  how 
many  pairs  of  gloves  will  $4  buy?  Ans.  8  pairs. 

44.  A.  B,  and  C  divide  $100  among  themselves,  B  takinji  S3  more 
than  A,  C  $4  more  than  B  ;  what  is  C's  share  ?  Ans.  $37. 

45.  A  man  would  put  30  gallons  of  mead  into  an  equal  number  of  1 
pint  and  2  pint  bottles  :  how  many  of  each  ?  Ans.  80. 

46.  A  merchant  puts  12  c\vt.  3  qrs.  12  Ibs.  of  tea  into  an  equal  num 
ber  of  5  lb.,  7  lb.,  and  12  Ib.  canisters  ;  how  many  of  each  ?     Ans.  60. 

47.  If  18  grs.  of  silver  make  a  thimble,  and  12  pwts  make  a  tea 
spoon,  how  many  of  each  can  be  made  from  15  oz.  6  pwts.  ? 

Ans.  24. 

48.  If  60  cents  be  divided  among  3  boys  so  that  the  first  has  3  cents 
as  often  as  the  second  has  5  and  the  third  7,  what  docs  each  receive  ? 

Ans.  12,  20,  and  28  cents. 

49.  A  gentleman  paid  $18*90  among  his  laborers,  to  each  boy  $-06, 
to  each  woman  $'08,  to  each  man  $'16;  there  were  three  women  for 
each  boy,  and  2  men  for  each  woman ;  how  many  men  were  there  ? 

Ans.  90. 

50.  A  man  paid  $82<50  for  a  sheep,  a  cow,  and  a  yoke  of  oxen ;  for 
the  cow  8  times,  for  the  oxen  24  times  as  much  as  for  the  sheep  ;  what 
did  he   pay  for  each?  Ans.  $2-50,  $20,  and  $60. 

51.  Three  merchants  accompanied  ;  A  furnished  •§•  of  the  capital,  B 
§,  and  C  the  rest ;  what  is  C's  share  of  $1250  gain  ?       Ans.  $281*25. 

52.  A  puts  in  $500,  B  $350,  and  C  120  yards  of  cloth;  they  gain 
$332<50,  of  which  C's  share  is  $120  ;  what  is  C's  cloth  worth  per  yard, 
and  what  is  A's  and  B's  share  of  the  gain  ? 

Ans.  C's  cloth  $4  per  yd.,  A's  share  $125,  B;s  do.  $87<50. 

53.  A,  B,  and  C  bought  a  farm,  of  which  the  profits  were  $580'80  a 
year ;  A  paid  towards  the  purchase  $5  as  often  as  B  paid  $7,  and  B  $4 
is  often  as  C  paid  $ri  ;  what  is  each  one's  share  of  the  gain  ? 

Ans.  A's  share  $129<066|,  B's  $180-693£,  C's  $271<04. 


IT  234.  MISCELLANEOUS   EXAMPLES.  297 

54.  A  gentleman  divided  his  fortune  among  his  sons,  giving  A  $9  as 
often  as  B  $5,  and  C  $3  as  often  as  B  $7  ;  C  received  $7442<10£  ;  what 
was  the  whole  estate  ?  Ans.  $56063-857f . 

55.  A  and  B  accompany;  A  put  in  $1200  Jan.  1st,  B  put  in  such  a 
sum,  April  1st,  that  he  had  half  the  profits  at  the  end  of  the  year ;  how 
much  did  B  put  in  ?  Ans.  $1600. 

56.  Three  horses,  belonging  to  3  men,  do  work  to  the  amount  of 
$26<45 ;  A  and  B's  horses  are  supposed  to  do  |  of  the  work,  A  and  C's 
Y$,  B  and  C's  £$,  on  which  supposition  the  owners  are  paid  propor 
tionally  ;  what  does  each  receive  ?     Ans.  A  $11<50,  B  $5<75,  C  $9<20. 

57.  A  gay  fellow  spent  f-  of  his  fortune,  after  which  he  gave  $7260 
for  a  commission,  and  continued  his  profusion  till  he  had  only  $2178 
left,  which   was  f  of  what  he  had  after  purchasing  his  commission  ; 
what  was  his  fortune?  Ans.  $18295<20. 

58.  A  younger  brother  received  £1560,  which  was  -fa  of  his  elder 
brother's  fortune,  and  5f  times  the  elder  brother's  fortune  was  |  of 
twice  as  much  as  the  father  was  worth  ;  what  was  he  worth  ? 

Ans.  £19165  14s.  3?d. 

59.  A  gentleman  left  his  son  a  fortune,  T5^  of  which  he  spent  in  3 
months ;  f  of  |-  of  the  remainder  lasted  him  9  months  longer,  when  he 
had  only  £537  left ;  what  was  the  sum  bequeathed  him  by  his  father? 

Ans.  £2082  18s.  2T2Td- 

60.  A  general,  placing  his  army  in  a  square,  had  231  men  left,  which 
number  was  not  enough  by  41  to  enable  him  to  add  another  to  each 
side  5  how  many  men  in  the  army  ?  Ans.  19000. 

61.  A  military  officer  placed  his  men  in  a  square:  being  reinforced 
by  three  times  his  number,  he  placed  the  whole  again  in  a  square  : 
again  being  reinforced  by  three  times  his  last  number,  he  placed  the 
whole  a  third  time  in  a  square,  which  had  40  men  on  each  side  j  how 
many  men  had  he  at  first  ?  Ans.  100. 

62.  Suppose  thai  a  man  stands  80  feet  from  a  steeple,  that  a  line  to 
him  from  the  top  of  the  steeple  is  100  feet  long,  and  that  the  spire  is 
three  times  as  high  as  the  steeple  •  what  is  the  length  of  a  line  reaching 
from  the  top  of  the  spire  to  the  man  ?  Ans.  197  feet  nearly. 

63.  Two  ships  sail  from  the  same  port ;  one  sails  directly  east  at  the 
rate  of  10  miles,  the  other  directly  south  at  the  rate  of  7£  miles  an  hour; 
how  far  are  they  apart  at  the  end  of  3  days  ?  Ans.  900  miles. 

64.  How  many  acres  in  a  square  field  measuring  70'71  rods  between 
the  opposite  corners?  Ans.  15§  acres. 

65.  Supposing  that  the  river  Po  is  1320  feet  wide  and  10  feet  deep 
and  runs  4  miles  an  hour ;  in  what  time  will  it  discharge  a  cubic  mile 
of  water  into  the  sea? 

NOTE.  —  A  linear  mile  is  5280  feet.  Ans.  22  days. 

66.  If  the  country  which  supplies  the  river  Po  with  water  be  380 
miles  long  and  120  broad,  and  the  whole  land  upon  the  surface  of  the 
earth  be  62,700,000  square  miles,  and  if  the  quantity  of  water  discharged 
by  the  rivers  intQ  the  sea  be  everywhere  proportional  to  the  extent  of 
land  by  which  the  rivers  are  supplied,  how  many  times  greater  than  the 
Po  will  the  whole  amount  of  the  rivers  be  ?  Ans.  1375  times. 

67.  Upon  the  same  supposition,  what  quantity  of  water,  altogether, 
will  be  discharged  by  all  the  rivers  into  the  sea  in  a  year  of  365  days  1 

Ans.  228124  cubic  miles. 


298  MEASUREMENT.  H"  235 

68.  If  the  proportion  of  sea  to  land  he  as  10£  to  5,  and  the  average 
depth  of  the  sea  be  1£  miles,  in  how  long  time,  if  the  sea  were  empty 
would  it  be  filled?  Ans.  8657  years  275  days. 

69.  If  a  cubic  foot  of  water  weigh  1000  oz.,  and  mercury  be  13£  times 
heavier  than  water,  and  the  hight  of  the  mercury  in  the  barometer 
(which  weighs  the  same  as  a  column  of  air  on  the  same  base  and  ex 
tending  to  the  top  of  the  atmosphere)  be  30  inches,  what  will  the  air 
weigh  on  a  square  foot?  —  on  a  square  mile?    What  will  the  whole 
atmosphere  weigh  ? 

Ans.,  in  order,  2109<375  Ibs.,  5880^000000  Ibs.,  11430122220000000000 
Ibs. 

70.  A  traveler  who  had  set  a  perfectly  accurate  watch  by  the  sun  in 
Boston,  71°  4'  W.  Ion.,  being  in  Detroit,  82°  58'  W.  Ion.,  3  days  after, 
was  surprised  to  find  it  wrong,  when  compared  with  the  sun  ;  was  it  too 
fast  or  too  slow?  how  much,  and  why  ? 

71.  A  building  fell  in  Portland,  Me.,  70°  20'  W.  Ion.,  at  9  o'clock,  A. 
M.,  and  in  3  minutes  the  intelligence  of  the  event  reached  St.  Louis, 
Mo.,  90°  15'  W.  Ion.,  by  magnetic  telegraph  •  when  was  it  known  at  St. 
Louis?  Ans.  At  43  m.  20  sec.  past  7  o'clock,  A.  M. 

72.  At  the  battle  of  Bunker  Hill  the  roar  of  cannon  was  distinctly 
heard  at  Hanover,  N.  H,  and  business  was  suspended  for  a  time  ;  in 
what  time  did  the  sound  pass,  the  distance  being  supposed  120  miles  ? 

NOTE.  —  Sound  moves  1142  feet  in  a  second.  Ans.  9  m.  14  sec.  -f% 

73.  Seeing  the  flash  of  a  rifle  in  the  evening,  it  was  8  seconds  before 
I  heard  the  report ;  what  was  the  distance  ?  Ans.  1  mi.  3856  ft. 

74.  A  man  in  view  on  a  hill  opposite  is  chopping,  at  the  rate  of  a 
blow  in  2  seconds  ;  I  saw  him  strike  4  blows  before  I  heard  the  first ; 
what  is  his  distance  from  me  ?  Ans.  1  mi.  1572  ft. 

75.  A  laborer  dug  a  cellar,  the  length  of  which  was  2  times  the  width, 
and  the  width  3  times  the  depth;  he  removed  144  cubic  yards  of  earth; 
what  was  the  length  ?  A?is.  36  feet. 

76.  A  owes  B  $750,  due  in  8  months ;  but  receiving  8300  ready 
money,  he  extends  the  time  of  paying  the  remainder,  so  that  B  shall 
lose  nothing;  when  must  it  be  paid?         Ans.  In  1  yr.  1  mo.  10  days. 

77.  The  sum  of  two  numbers  is  266§,  and  the  product  of  the  greater 
multiplied  by  3,  equals  the  product  of  the  less  multiplied  by  5  ;  what  are 
the  numbers?  Ans.  100,  and  166§. 

78.  A  park  10  rods  square  is  surrounded  by  a  walk  which  occupies 
of  the  whole  park  ;  what  is  its  width  ?  Ans.  8  ft.  3  in. 

79.  A,  B  and  C  commence  trade  with  $3053*25,  and  gain  $610<65; 
A.'s  stock  -j-  B's,  is  to  B's  -4-  C's,  as  5  to  7 ;  and  C's  stock  —  B's,  is  to 
C's  -4-  B's,  as  1  to  7  ;  what  is  each  one's  part  of  the  gain  ? 

Ans.  A's  gain  $135<70,  B's  $203<55,  C's  $271<40. 


MEASUREMENT    OF    SURFACES. 

IT  235.    ToJiTid  the  area  of  a  parallelogram,  multiply  the 
length  by  the  shortest  distance  between  the  sides. 


11235. 


MEASUREMENT 


299 


NOTE  1.  —  A  parallelc  gram  has  its  opposite 
sides  equal,  but  its  adjacent  sides  unequal,  like 
the  figure  A  B  C  D,  or  E  F  C  D.    The  former  is 
called  a  rectangle,  see  IT  48.  The  second  is  called 
p       8     a  rhomboid,  and  is  equal  in  size  to  the  first. 

1.  What  are  the  superficial  content^  if  an  oblique  angled  piece  of 
ground,  measuring  80  rods  in  length  aru.  20  rods  in  a  perpendicular 
line  between  its  sides  ?  Ans:  1600  sq.  rods. 

NOTE  2.  —  To  find  the  contents  c  ?  rhombus,  which,  like 
the  annexed  figure,  has  its  sides  eq-'Bt,  but  its  angles  not 
right  angles  ;  multiply  the  length  of  one  side  by  the  shortest 
distance  to  the  side  opposite. 

To  find  the  area  of  a  trapezoid,  multiply  half  the  sum  of 
the  parallel  sides  by  the  shortest  distance  between  them. 

NOTE  3.  — A  trapezoid  is  a  figure,  like  the  one  in  the  an 
nexed  diagram,  bounded  by  four  straight  lines,  only  two  of 
which  are  parallel. 

2.  What  is  the  area  of  a  piece  of  ground  in  the  form  of  a  trape 
zoid,  one  of  whose  parallel  sides  is  8  rods,  the  other  12  rods,  and  the 
perpendicular  distance  between  them  16  rods  ? 

84^-SL  x  16  =  160    sq.  rods,  Ans. 

3.  How  many  square  feet  in"a  board  16  feet  long,  1*8  feet  wide  at 
one  end,  and  1*3  at  the  other1?  Ans.  24'8  feet. 

To  find  the  area  of  a  triangle,  multiply  the  base  by  half 
the  altitude. 


NOTE  4.  —The  figure  ABC  is  a  trian 
gle,  of  which  the  side  A  B  is  the  base,  D  C 
the  altitude.  The  triangle  is  evidently 
half  the  parallelogram  A  B  C  F,  the  area 
of  which  equals  A~B  X  D  C. 


A  D 

4.  The  base  of  a  triangle  is  30  rods,  and  the  perpendicular  10 
rods;  what  is  the  area?  Ans.  150  rods. 

5.  If  the  contents  are  600  rods,  and  the  base  75  rods,  what  is  the 
altitude  ?  Ans.  16  rods. 

6.  Required  the  base,  the  area  being  400,  the  altitude  40  rods 

Ans.  20  rodfc 

7.  How  many  square  feet  in  a  board  18  feet  long,  l£  feet  wide  at 
one  end,  and  running  to  a  point  at  the  other?  Ans.  13£  feet. 

To  find  the  circumference  of  a  circle  when  the  diameter  is 
knoivn,  multiply  the  diameter  by  3^-,  or  accurately  by  3' 14159. 

To  find  the  area,  multiply  the  circumference  by  one  fourth 
of  the  dimeter  •  or  multiply  the  square  of  the  diameter  by 
'7S54. 


300  MEASUREMENT.  11236 

NOTE  5.  —  The  principles  on  which  the  above  ru'es  are  founded,  as  well  aa 
those  for  the  measurement  of  many  figures,  will  be  understood  by  a  geometrical 
demonstration,  and  the  pupil  must  take  them  without  a  demonstration  till  he 
may  study  that  interesting  science. 

8.  What  is  the  circumference  of  a  circular  pond,  the  diameter  of 
which  is  147  feet?     What  is  the  area? 

Ans.  to  the  last,  16971-fo-  feet. 

9.  If  the  circumference  be  22  feet,  what  is  the  diameter? 

Ans.  7  feet. 

10.  If  the  diameter  of  the  earth  is  7911  miles,  what  is  the  circum 
ference?  Ans.  24853  miles. 

11.  How  many  square  inches  of  leather  will  cover  a  ball  3i  inches 
in  diameter  ? 

NOTE  6.  —  The  area  of  a  ball  is  4  times  the  area  of  a  circle  having  the  same 
diameter.  Ans.  38i  square  inches. 

12.  How  many  square  miles  on  the  earth's  surface? 

Ans.  196,612,083. 


Measurement  of  Solids. 


Tl  "2S6.      NOTE  1  .  —  The  general 


ral  principle  for  finding  the  contents  of 

solid  bodies  is  to  multiply  the  length  by  the  breadth,  and  the  product  by  the 
thickness,  but  the  rule  applies  directly  only  to  the  cube  or  right  prism,  being 
subject  to  modifications  as  applied  to  solid  figures  of  other  forms.  See  IF  51. 

1.  How  many  solid  inches  in  a  globe  7  inches  in  diameter  ? 

NOTE  2.  —  The  solid  contents  of  a  globe  are  found  by  multiplying  the  area 
of  its  surface  by  ^  pan  of  its  diameter,  or  the  cube  of  its  diameter  by  '5236. 

Ans.  179|  solid  inches. 

2.  What  number  of  cubic  miles  in  the  earth  ? 

Ans.  259,233,031,  435$. 

3.  What  are  the  solid  contents  of  a  log  20  feet  long,  of  uniform 
size,  the  diameter  of  each  end  being  2  feet  ? 

NOTE  3.  —  A  figure  like  the  above  is  called  a  cylinder.  To  find  the  solid 
contents,  we  find  the  area  of  one  end  by  a  foregoing  rule,  and  multiply  the  area 
thus  found  by  the  length. 

Ans.  62'83  -f-  cu.  ft. 

4.  A  bushel  measure  is  18'5  inches  in  diameter,  and  8  inches  deep; 
•low  nnany  cubic  inches,  does  it  contain?  Ans.  2150'4-{-. 

N-.TE4.  —  Solids  having  bases  bounded  by  straight  lines,  and  decreasing 
unnormly  till  they  come  to  a  point,  are  called  pyramids.  Solids  which  thus 
decrease,  with  circular  bases,  are  called  cones.  Pyramids  and  cones  are  just 
one  third  as  large  as  cylinders,  of  \vhich  the  area  of  each  end  is  equal  to  the 
area  of  the  bases  of  these  solids.  Hence,  if  we  multiply  the  area  of  the  base 
by  the  hight,  and  divide  the  product  by  3,  the  quotient  will  be  the  solid  con 
tents. 

5.  What  are  the  solid  contents  of  a  pyramid,  the  base  of  which  ia 
4  feet  square,  and  the  perpendicular  hight  9  feet? 

Ans.  48  solid  feet. 


II 237, 238.  MEASUREMENT.  301 

6.  What  are  the  solid  contents  of  a  cone,  the  hight  of  which  is  27 
feet,  and  the  diameter  of  the  base  is  7  feet  ?      Ans.  346£  solid  feet. 

7.  What  are  the  solid  contents  of  a  stick  of  timber  18  feet  long, 
one  end  of  which  is  9  inches  square  and  the  other  end  4  inches  square, 
uniformly  diminishing  throughout  its  whole  length? 

NOTE  5.  —  Such  a  figure  is  called  the  frustum  of  a  pyramid,  and  the  solid 
contents  are  found  by  adding  to  the  areas  of  the  ends  the  square  root  of  their 
product,  and  multiplying  the  sum  by  one  third  of  the  hight.  The  pupil  must 
notice  that  the  diameters  are  expressed  in  inches,  while  the  length  is  in  feet. 

Ans.  5  solid  feet,  936  solid  inches. 

8.  What  are  the  solid  contents  of  a  round  log  of  wood,  36  feet 
long,  1'6  feet  in  diameter  at  one  end,  and  diminishing  gradually  to  a 
diameter  of  '9  of  a  foot  at  the  other? 

NOTE  6.  —  Such  a  figure  is  called  the  frustum  of  a  cone,  and  the  solid  con 
tents  are  found  by  adding  to  the  squares  of  the  two  diameters  the  square  root 
of  the  product  of  those  squares,  multiplying  the  sum  by  '7854,  and  the  result 
ing  product  by  one  third  of  the  length. 

Ans.  45<  333 -f  solid  feet. 


Gauging*,  or  Measuring  Casks. 

^T  2367.  1.  Plow  many  gallons  of  wine  will  a  cask  contain,  the 
head  diameter  of  which  is  25  inches,  and  the  bung  diameter  3.1  inches, 
and  the  length  36  inches?  How  many  beer  gallons  ? 

NOTE.  —  Add  to  the  head  diameter  2  thirds,  or  if  the  staves  curve  but 
slightly,  6  tenths  of  the  difference  between  the  head  and  bung  diameters,  the 
sum  will  be  the  average  diameter.  The  cask  will  then  be  reduced  to  a  cylin 
der,  the  contents  of  which  may  be  found  by  a  foregoing  rule,  in  solid  inches. 
The  solid  inches  may  be  divided  by  231,  (IT  114,)  to  find  the  number  of  wine 
gallons  which  the  cask  will  contain,  and  by  282,  (IT  115,)  to  find  the  number 

[1°nS'     Ans.  102<93  +  wine  gallons,  84'32  +  beer  gallons. 
2.    How  many  wine  gallons  in  a  cask,  the  bung  diameter  of  which 
is  36  inches,  the  head  diameter  27  inches,  and  the  length  45  inches? 

Ans.  166'617. 


Mechanical  Powers. 

*f[  S38,  1.  A  lever  is  10  feet  long,  and  the  fulcrum,  or  prop, 
on  which  it  turns  is  2  feet  from  one  end  ;  how  many  pounds  weight 
at  the  short  end  will  be  balanced  by  42  pounds  at  the  other  end  ? 

NOTE  i.  —  In  turniug  round  the  prop,  the  long  end  will  evidently  pass  over 
a  space  of  3  inches,  v.  hile  the  short  cud  passes  over  a  space  of  2  inches.  Now, 
it  is  a  fundamental  principle  in  mechanics,  that  the  weight  and  power  will 
exactly  balance  each  other,  when  they  are  inversely  as  the  spaces  they  pass 
over.  Hence,  in  this  example,  2  pounds,  8  feet  from  the  prop,  will  balance  8 
pounds  2  feet  from  the  prop  ;  therefore,  if  we  divide  the  distance  of  the  POWER 
from  the  prop  by  the  distance  of  the  WEIGHT  from  the  prop,  the  quotient  will 
always  express  the  ratio  of  the  WEIGHT  to  the  POWER;  J-=4,  that  is,  the 
weight  will  be  4  times  as  much  as  the  power. 

42X4=168.  Ans.  168  Ibs. 

26 


302  MEASUREMENT.  IT  238. 

2.  Supposing  the  lever  as  above,  ivhat  power  would  it  require  to 
raise  1000  pounds?  Ans.  i^.  =  250  pounds. 

3.  If  the  weight  to  be  raised  be  5  times  as  much  as  the  power  to 
be  applied,  and  the  distance  of  the  weight  from  the  prop  be  4  feet, 
how  far  from  the  prop  must  the  power  be  applied  ?      Ans.  20  feet. 

4.  If  the  greater  distance  be  40  feet,  and  the  less  £  of  a  foot,  and 
the  power  175  pounds,  what  is  the  weight?       Ans.  14000  pounds. 

5.  Two  men  carry  a  kettle,  weighing  200  pounds  ;  the  kettle  is 
suspended  on  a  pole,  the  bale  being  2  feet  6  inches  from  the  hands  of 
one,  and  3  feet  4  inches  from  the  hands  of  the  other ;  how  many 

pounds  does  each  bear?  (  114-?-  pounds. 

Ans.  < 

(    85^-  pounds. 

6.  There  is  a  windlass,  the  wheel  of  which  is  60  inches  in  diame 
ter,  a^d  the  axis,  around  which  the  rope  coils,  is  6  inches  in  diameter ; 
now  many  pounds  on  the  axle  will  be  balanced  by  240  pounds  at  the 
.vheei  ? 

NOTE  2.  —  The  spaces  passed  over  are  as  the  diameters,  or  the  circumfer 
ences;  therefore,  -^  —  10,  ratio. 

Ans.  2400  pounds. 

7.  If  the  diameter  of  the  wheel  be  60  inches,  what  must  be  the 
diameter  of  the  axle,  that  the  ratio  of  the  weight  to  the  power  may  be 
10  to  1  ?  Ans.  6  inches. 

NOTE  3.  —  This  calculation  is  on  the  supposition,  that  there  is  no  fraction, 
for  which  it  is  usual  to  add  £  to  the  power  which  is  to  work  the  machine. 

8.  There  is  a  screw,  the  threads  of  which  are  1  inch  asunder  ;  if  it 
is  turned  by  a  lever  5  feet,  =60  inches,  long,  what  is  the  ratio  of  the 
weight  to  the  power? 

NOTE  4.  —  The  power  applied  at  the  end  oi  the  lever  will  describe  the  cir 
cumference  of  a  circle  GO  X  2  =  120  inches  in  diameter,  while  the  weight  is 
raised  i  inch  ;  therefore,  the  ratio  will  be  found  by  dividing  the  circumference 
of  a  circle,  ichose  diameter  is  twice  the  length  of  the  lever,  by  the  distance  be 
tween  the  threads  of  the  screw. 

Q-.-V    £ 

120  X  3J-  =  377|  circumference,  and *  =  377f,  ratio,  Ans. 

9.  There  is  a  screw,  whose  threads  are  ^  of  an  inch  asunder  ;  if  it 
be  turned  by  a  lever  10  feet  long,  what  weight  will  be  balanced  by 
120  pounds  power  ?  Ans.  362057|  pounds. 

10.  There  is  a  machine,  in  which  the  power  moves  over  10  feet, 
while  the  weight  is  raised  1  inch  ;  what  is  the  power  of  that  machine, 
that  is,  what  is  the  ratio  of  the  weight  to  the  power?        Ans.  120. 

11.  A  man  put  20  apples  into  a  wine  gallon  measure,  which  was 
afterwards  filled  by  pouring  in  1  quart  of  water  ;  required  the  con 
tents  of  the  apples  in  cubic  inches.  Ans.  173^  inches. 

12.  A  rough  stone  was  put  into  a  vessel,  whose  capacity  was  14 
wine  quarts,  which  was  afterwards  filled  with  2^  quarts  of  water ; 
what  was  the  cubic  content  of  the  stone?  -4ns.  664  J  inches. 

NOTE  5.  —  For  a  more  full  consideration  of  the  foregoing  subjects,  the  pupil 
is  referred  to  the  forthcoming  treatise  on  Mensuration  in  connection  witn  the 
"series." 


FORMS  OP  NOTES,  ETC.  303 

FORMS    OF    NOTES,    RECEIPTS,    AND 
ORDERS. 

Notes. 

No.  I. 

Keene,  Sept.  17.  1846. 

For  value  received,  I  promise  to  pay  to  OLIVER  BOUNTIFUL,  or  order,  sixty- 
Jiree  dollars,  fifty-four  ceuts,  on  demand,  with  interest  after  three  mouths. 
Attest,  TIMOTHY  TESTIMONY.  WILLIAM  TRUSTY. 

No.  II. 

Ludlow,  Sept.  17,  1846. 

For  value  received,  I  promise  to  pay  to  O.  R.,  or  bearer,  dollars 

?ents,  three  mouths  after  date.  PETER  PENCIL. 

No.  III. 
By  two  persons. 

Yates,  Sept.  17,  1846. 
For  value  received,  we,  jointly  and  severally,  promise  to  pay  to  C.  D.,  or 

order, dollars cents,  on  demand,  with  interest. 

Attest,  PETER  SAXE.  ALDEN  FAITHFUL. 

JAMES  FAIRFACB. 


Receipts. 

Boston,  Sept.  19,  184G. 
Received  from  Mr.  DURANCE  ADLF.Y  ten  dollars  in  full  of  all  account^: 


ORVAND  CONSTANCE. 


Newark,  Sept.  19,  1846. 

Received  of  Mr.  ORVAND  CONSTANCE  five  dollars  in  full  of  all  accounts. 

DURANCE  ADLEY. 


Receipt  for  Money  received  on  a  Note. 

Rochester,  Sept.  19,  rS46. 

Received  of  Mr.  SIMPSON  EASTEY  (by  the  hand  of  TITUS  TRUSTY)  sixteen 
dollars  twenty-five  cents,  which  is  endorsed  on  his  note  of  June  3,  1846. 

PETER  CHEERFUL. 


A  Receipt  for  Money  received  on  Account. 

Hancock,  Sept.  19,  1846. 
Received  of  Mr.  OR  LAND  LANDIKE  fifty  dollars  on  account. 

ELDRO  SLACKLEY. 


Receipt  for  Money  received  for  another  Person. 

T 

ELI  TRUMAN. 


Salem,  August  10,  1846. 
Received  from  P.  C.  one  hundred  dollars  for  account  of  J.  B. 


Receipt  for  Interest  due  on  a  Note. 

Amherst,  July  6,  1846. 
Received  of  I.  S.  thirty  dollars,  in  full  of  one  year's  interest  of  $500,  due  to 

me  ou  the day  of last,  on  note  from  the  said  I.  S. 

SOLOMON  GRAY. 


304  FORMS  OF  BILLS. 

Receipt  for  Money  paid  before  it  becomes  due. 

Hillsborough,  May  3,  1846. 

Received  of  T.  Z.  ninety  dollars,  advanced  in  full  for  one  year's  rent  of  my 
farm,  leased  to  the  said  T.  Z.,  ending  the  first  day  of  April  next,  1^47. 

HONESTUS  JAMES. 

NOTE.  — There  is  a  distinction  between  receipts  given  in  full  of  all  accounts, 
and  others  in  full  of  all  demands.  The  former  cut  off  accounts  only;  the  lat 
ter  cut  off  not  only  accounts,  but  all  obligations  and  right  of  action. 


Orders. 

Utica,  Sept.  9,  1846. 

Mr.  STEPHEN  BURGESS.  For  value  received,  pay  to  A.  B.,  or  order,  ten 
dollars,  and  place  the  same  to  my  account.  SAMUEL  SKINNER. 

Pittsburgh,  Sept.  9,  1846. 

Mr.  JAMES  ROBOTTOM.  Please  to  deliver  to  Mr.  L.  D.  such  goods  as  he 
may  call  for,  not  exceeding  the  sum  of  twenty-five  dollars,  and  place  the  same 
to  me  account  of  your  humble  servant,  .  NICHOLAS  REUBENS. 


FORMS    OF    BILLS. 

"  Before  you  build,  sit  clown  and  count  the  cost." 

S&neon  Thrifty  built  a  house  for  Thomas  Paywell,  according  to  a  plan 
agreed  upon  between  them,  for  the  sum  of  $1500.  The  cellar  of  the  house  is 
24  by  28  feet,  and  is  dug  4  feet  deep  below  the  top  of  the  ground.  The  cellar 
walls  are  7  feet  high.  There  is  a  wing  at  one  end  of  the  main  building,  20  by 
24  feet,  which  is  underpinned  with  a  wall  3  feet  high.  As  one  side  of  the  wing 
joins  the  main  building,  for  the  underpinning  of  it  but  3  walls  are  required,  one 
24,  and  two  20  feet  long  on  the  outside.  The  walls  of  the  cellar,  and  the  un 
derpinning  of  the  wing  are  li  feet  thick.  To  the  cellar  there  is  a  door  4  by  7 
feet,  and  2  windows,  each  2  by  2*  feet.  Simeon  Thrifty,  wishing  to  know 
how  much  it  cost  him  to  build  the  house,  kept  an  accurate  account  of  all  the 
materials  used,  the  labor  employed,  and  the  cost  of  each.  The  following  are 
his  bills:  — 


Bill  of  Timber. 
part, e: 

sills  to  upright  part, "     28 


6  sticks  for  posts  to  upright  part, each  14  ft.  long,  and  4  by  10  in. 

6         '  "         wing, "11  4  "    10  " 

rnt 


and  sleepers  to  upright  and 


24  7  "  8  " 

plates  and  sido  girts  to  upright,  "     28  6  "  7  " 
"      and  beams  to  upright  and 

wing, "24  6  "  7  " 

2  "         eave  gutters, "30  6  "  10  " 

3  "                     "           "20  6  "  10 

52  rafters, "15  3  "  4 

96  studs, "12  2  "  4 

175  partition  plauks, *     10  2  "  4 

96  scantlings, "12  4  "  4 

10        <:           for  braces, "12  4  "  4 

75  joists,            "     14  2  "  7 

30     :<                                                              .    ..."      10              "  8  «  7 


FORMS  OF  BILLS. 


305 


Bill  of  Lumber. 


800  ft.  best  quality  pine,  for  best  doors,  and  other  nice  joiner  work. 
10000  "  '       "      " 


common 


,  . 

door  and  window  casings,  stairs,  base  boards 

common  doors,  to-'.,  &c. 
3850  "  white  wood  siding. 
2160  "  bass        "     flooring. 
150C      roof  boards. 
COOO  "  lath. 
26  bunches  shingles,  500  in  each  bunch. 


Bill  of  Materials  for  Windows. 
Sash  and  glass  for  14  windows  of  24  panes  each,  7  jy  9  inches. 


30  Ibs.  putty. 


4 

2         " 
1  window, 
i          « 


d     / 


Hardware  Bill. 


4  casks  nails,  100  Ibs.  each. 
22  pairs  3  inch  door  hinges,  with  screws. 

2      "      4     "  " 

20  door  handles,  " 

2  outside  door  knobs  and  locks. 

3  cupboard  fastenings. 

63  ft.  tin  eave  conductors,  including  4  elbows. 
3  stove-pipe  crocks. 
3        "  thimbles. 

3  ft.  tin  pipe  for  sink-spout. 
20  window  springs. 

4  papers  1  inch  brads. 

Bill  of  Materials  for  Chimneys  and  Plastering. 

1  COO  bricks.  27  loads  sand. 

200  bush.  lime.  10  bush.  hair. 

Bill  of  Prices  of  Materials. 

Stone  for  cellar  walls  and  underpinning,    ....  $     '25  per  perch. 
All  the  timber  except  the  eave  gutters  reduced  to 

board  measure,  that  is,  1  inch  thick,   ...  10'        "  M. 

Eave  gutters, 15£        " 

Pine  lumber,  best  quality, 20'        "  " 

"        "        common, 10' 

Siding,  flooring,  and  roof  boards, 10'         '  " 

Lath,    .   .   . 5'        '  « 

Shingles, 1'50     '  bunch. 

Window  sash, '03     '  pane. 

"        glass, 2'50     '  box  of  1 14  pane* 

Putty, '07     Mb. 

Nails, '05i  '  "  _ 

3  inch  door  butts,  with  screws, '12i  "  pair. 

4  «  "  "  '15   "      " 

Door  handles, '12J  each. 

Outside  door  knobs*and  locks, 1'50 

Cupboard  fastenings, '12J    " 

Eave  conductors, '12i  per  ft. 

Extra  for  elbows, '06i  each. 

Stove-pipe  crocks, '37i     " 

"          thimbles, '12i     " 

26* 


306 


FORMS  OF  BILLS. 


Sink  spout, $    '44. 

Window  springs, '06i  each. 

Brads, '10  per  paper. 

Bricks, 10'        "    M. 

Lime, '12*  "    bush. 

Sand, I'OO    "    load. 

Hair, '25    "    bush. 


Bill  of  Prices  of  Labor. 


Digging  cellar, 

6  stone  masons,    6  days  each, , 

2  carpenters,        12  , 

3  joiners,  40         "  

Painting  and  glazing, , 

Furring  ready  for  lathing, 

Lathing, 

2  plasterers,  7  days  each, 

3  brick-layers,  1  day  "       

Team  and  hired  man,  4  months  of  26  days  each, .   .   . 

Simeon  Thrifty  commenced  the  house  on  the  1st  day  of  May,  and  completed 
it  on  the  3d  day  of  Sept. ;  allowing  him  $1'50  per  week  for  his  board,  how 
much  did  he  get  for  his  own  labor  1  Ans.  $281 '64-. 


I      C18  per  cu.  yd. 

2'00    "    day. 

1'50    "      "" 

1'75    "      " 
lOO'OO. 
12'00. 
IS'OO. 

2 '50  per  day. 

2 '75        " 

2[00        " 


75-j§j-  perches  stone,  .    . 
600  ft.  cave  gutters,    .   . 
8287  "   other  timber,   .   . 
800  "  best  pine  lumber, 
17510  "  other  lumber,   .   . 

9  IS'SS-jZj- 
9'00 
82[87 
16(00 
175'10 
30(00 
39(00 
13(68 
lO'OO 
2'10 
22'00 
3(05 
2(50 
3'00 
'37* 
7'87* 
'25 

i'm 

'37* 
'44 
1'25 

Brads,    

1600  bricks,  

.    .        9'60 

.      25'00 

27  loads  sand  ... 

.    .      27'00 

10  bush,  hair,  .   .   . 

.    .        2<50 

.    .        4'48 

26  bunches  shingles,     . 

Laying  stone  work,    . 

.  .     eo'oo 

.    .      36(00 

4  boxes  window  glass,    . 
30  Ibs   putty  ...       .   . 

Joiners'            "        .   . 
Painting  and  glazing, 

.    .    210'00 
.    .    lOO'OO 
.    .       12'00 

.    .       15'00 

Plastering,    

.    .      35'00 

Outside  door  knobs  &  locks. 
Cupboard  fastenings, 
Tin  eave  conductors,  . 
Extra  on  elbows,     .   . 
Stove-pipe  crocks,  .   . 
"          thimbles,  . 
Sink  spout,   

Brick-laying         . 

.    .        7'50 

Team  and  hired  man, 

.   .    208'00 
.   .      27'00 

Amount,  .  .  . 
Errors  excepted, 

.    S1218'35^ 

j.  n  *. 

Window  springs,     ... 

SCHOOL   BOOK   DEPOS   TORY. 
PHILLIPS  &  SAMPSON, 

PUBLISHERS,   BOOKSELLERS,  AND   STATIONERS, 
110  WASHINGTON  STREET, 

(Up  Stairs,) 
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and  others  purchasing  School  Books,  are  respectfully 
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Particular  attention  is  paid  to  furnishing  all  the 
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Also,  a  very  extensive  assortment  of  Standard,  Theo 
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Paper,  Quills,  Ink,  Slate-s,  &c.  of  every  variety. 

Orders  solicited  and  promptly  attended  to. 


THOMAS  SHERWIN,  A  M.,  Principal  of  the  English  High  School,  Boston 

BARNUM  FIELD,  "          "        Franklin  "  " 

SAMUEL  S.  GREENE,  Principal  of  the  Phillips  School,  Boston. 

JOSHUA  BATES,  JR.,         "          "      Brimmer    " 

JOSIAH  A.  STEARNS,        "  "      Mather 

ISAAC  F.  SHEPARD,          "          "      Otis'          " 

GEORGE  B.  HYDE,  "      Dwight      "  " 

J.  D.  PHILBRICK,  Principal  of  the  Mathematical  Depar't  of  May  hew  School,  Boston. 

WILLIAM  A.  SHEPARD,  "  "  "  "  "  " 

D.  P.  PAGE,  "  New  York  State  Normal  School,  Albany. 

P.  H.  SWEETSEB,  Harvard  School,  Charlestown. 

ELBRIDGE  SMITH,  "  Classical  and  English  High  School,  Worcester. 

C.  C.  DEANE,  "  English  High  School,  Newburyport. 

HON.  FRANCIS  DWIGHT,  late  of  Albany,  N.  Y. 

JAMES  D.  BATCHELDER,  Principal  of  Grammar  School,  Marblehead. 

CHARLES  EDWARDS, 

Academy  at  Concord,  Mass. 

Union  Seminary,  Fairhaven. 

Brown    School,    Salem. 

Hacker 

Pickering 

Phillips 

Epes 

Fish 

Salstonstall 

Grammar  School,  Lynn. 


CHARLES  W.  GOODNOW, 
ALONZO  TRIPP, 
J.  B.  FAIRFIELD, 

D.  P.  GALLOUP, 
ALBERT  LACKEY, 
JOSEPH  WILLIAMS, 
CHARLES  NORTHEND, 
OLIVER  CARLTON, 
EDWIN  JOCELYN, 
JACOB  BATCHELDER,  Jr., 
WILLIAM  S.  WILLIAMS, 
AMASA  DAVENPORT, 
ELWELL  WOODBURY, 
WILLIAM  T.  ADAMS, 
JOHN  CAPEN, 

ISAAC  SWAN, 

GEORGE  NEWCOMB,  Teacher  of  Quincy  Grammar  School. 

E.  WYMAN,  Principal  of  the  English  and  Classical  High  School,  St.  Louis,  Mo. 
CHARLES  A.  LORD,  A.  M.,  late  Professor  in  Marion  College,  Ohio. 

Z.  GROVER,  Principal  of  Prospect  street  School, 


Teachers  of  the  Dorchester  Grammar  Schools. 


L.  B.  NICHOLS, 
J.  D.  GIDDINGS, 
C.  FARNUM, 
AMOS  PERRY, 
S.  S.  ASHLEY, 
C.  T.  KEITH, 
EBENEZER  HARVEY, 
BENJAMIN  EVANS, 
CYRUS  BARTLETT, 


Arnold 

Fountain 

Elam 

Summer 

Meeting 

Benefit 


Providence,  R.  L 


Principals  of  Grammar  Schools,  New  Bedford. 


COMMON  SCHOOL  ALGEBRA,  by  Thomas  Sherwin,  A.  M.,  Principal  of  the 
English  High  School,  Boston;  author  of  Elementary  Treatise  on  Algebra,  <5cc. 

This  Algebra  is  now  used  as  the  text  book  in  the  Public  Schools  of  Boston,  Rox- 
bury,  Salem,  &c. ;  also,  in  the  Phillips'  Academy,  Andover;  Bradford  Academy, 
Bradford ;  Windsor  Academy,  Windsor,  Vt. ;  Fitchburg  Academy,  and  numerous 
other  places  ;  and  has  been  recommended  by  the  Superintendents  of  Public  Schools 
for  the  whole  state  of  Rhode  Island.  A  large  number  of  testimonials  from  our  best 
practical  teachers  are  in  the  hands  of  the  publishers,  from  which  they  would  take  the 
following : 

PHILLIPS  SCHOOL,  March  13,  1846 

MR.  SHERWIN,  -Sir :  I  have  examined  your  "  Common  School  Algebra,"  and  believe 
it  better  adapted  to  the  wants  of  the  beginner  than  any  other  I  have  seen.  The  plan 
of  introducing  preliminary  exercises,  to  acquaint  the  pupil  with  the  use  of  letters,  is 
a  happy  one.  By  the  aid  of  these  introductory  lessons,  the  transition  from  arithme 
tic  to  algebra  is  made  so  easy  that  a  child  of  ordinary  capacity  would  meet  with  little 
difficulty  in  determining  the  use  of  x,  y,  and  z,  even  without  a  teacher.  I  hope  to 
see  the  book  extensively  used,  in  that  class  of  schools  for  which  it  was  prepared. 

SAMUEL  S.  GREENE, 
Grammar  Master  of  the  Phillips  School 

ROXBCRY,  February  5,  1846. 

MR.  SHERWIN,  —  Dear  Sir:  I  have  been  highly  gratified  by  the  examination  of  your 
"  Common  School  Algebra."  The  principles  of  the  science  are  unfolded  and  explained 
with  great  perspicuity  and  simplicity.  I  think  your  reasonings  and  illustrations  are 


43 


peculiarly  happy  and  appropriate ;  and,  on  the  whole,  I  consider  it  superior  to  any 
work  of  the  kind  that  I  have  ever  seen.    With  much  esteem, 
Your  obe'lieat  servant, 
LEVI  REED. 

ROXBUKY,  February  5,  1846. 

I  FCLLY  concur  in  the  opinion  of  Mr.  Reed  in  regard  to  Mr.  Sherwin's  "  Common 
School  Algebra."  I  have  introduced  it  into  my  school,  and  shall  esteem  it  a  privilege 
to  recommend  its  use  whenever  an  opportunity  presents. 

JEREMIAH  PLYMPTON. 

MAYHEW  SCHOOL,  Boston,  February  12,  1846. 

THOMAS  SHERWIN,  Esq,.,— Dear  Sir  :  I  have  examined  the  "  Common  School  Alge 
bra,"  which  you  sent  me,  and  think  it  better  adapted  to  the  wants  of  the  schoolroom 
than  any  other  book,  upon  the  same  subject,  with  which  I  am  acquainted. 

Yours  very  truly,       WILLIAM  D.  SWAN. 


BOSTON,  December  8,  1845. 

THOMAS  SHERWIH.  ESQ,.,  —  My  Dear  Sir :  I  have  examined  your  "  Common  School 
Algebra,"  which  you  kindly  sent  me,  with  great  satisfaction.  It  seems  to  be  a  very 
natural  and  lucid  development  of  the  principles  of  computation  by  algebraic  symbols, 
and  makes  the  .whole  subject  appear  much  more  practical  than  it  does  in  most  works 
of  this  description.  Indeed,  I  think  you  have  succeeded  remarkably  in  removing 
what  you  justly  call  the  "great  difficulty  in  the  study  of  algebra,"  by  helping  the 
pupil  to  "a  clear  comprehension  of  the  earliest  steps."  The  work  throughout  bears 
the  marks  of  an  experience  in  teaching,  as  well  as  theoretical  familiarity  with  the 
science  elucidated.  Accept  of  my  very  sincere  thanks,  and  assurance  of  the  truest 
esteem  and  affection.  H.  W1NSLOW. 

CENTRAL  PLACE,  Boston,  February  4,  1846. 

THOMAS  SHERWIN,  ESQ,.,— My  Dear  Sir :  I  have  examined  your  "  Common  School 
Algebra  "  with  much  pleasure.  It  seems  to  me  so  admirably  adapted  for  a  first  book 
in  the  science,  that  I  shall  adopt  it  for  my  next  class.  I  like  it,  because  it  appears 
in  every  part  to  be  the  work  of  a  practical  teacher,  who  has  observed  the  wants  of 
his  pupils,  and  skilfully  contrived  to  meet  them. 

Very  truly  yours. 
SOLOMON  ADAMS. 

BRIMMER  SCHOOL,  January  31st,  1846. 

THOMAS  SHERWIN,  ESQ.,  —  Dear  Sir:  I  have  carefully  examined  your  "Common 
School  Algebra,"  and  I  take  pleasure  in  expressing  to  you  my  opinion  of  its  merits. 
I  believe  it  to  be  the  best  elementary  treatise  upon  Algebra  that  has  ever  been  pub 
lished  for  the  use  of  our  schools.  Strictly  scientific,  it  is  yet  entirely  adapted  to  the 
capacity  of  that  class  of  scholars  for  whom  it  was  designed  ;  and  it  combines  very 
just  proportions  of  simplicity  and  strengtn.  The  selection  of  subjects  is  sufficiently 
extensive  to  render  it  useful  to  those  who  are  desirous  of  pursuing  a  higher  course 
of  mathematical  study,  without  being  too  severe  for  the  comprehension  of  young 
pupils  ;  and  the  careful  and  judicious  arrangement  of  the  progressive  exercises  cannot 
fail  to  reduce  it  to  the  level  of  less  than  medium  ability.  As  a  practical  teacher,  I 
heartily  thank  you  for  your  efforts  to  facilitate  the  acquisition  of  this  important  sci 
ence.  Yours  with  respect, 

WILLIAM  A.  SHEPARD. 

MR.  SHERWIN'S  "Common  School  Algebra  "  I  regard  as  a  work  of  superior  merit, 
and  consider  it  decidedly  preferable  to  any  other  elementary  text-book,  upon  the  same 
subject,  with  which  I  am  acquainted.  Its  chief  peculiarity,  if  not  its  highest  excel 
lence,  is  to  be  found  in  the  "  preliminary  exercises,"  which  open  a  new  and  delight 
ful  avenue  to  this  branch  of  mathematical  science.  They  are  admirably  calculated  to 
communicate  to  the  learner  the  first  principles  of  Algebra  in  the  most  simple  and 
intelligible  manner.  Having  made  use  of  them  in  my  class  last  year,  I  can  speak  of 
their  excellence  with  confidence.  In  a  ward,  the  book  is  just  the  thin^  we  need  in 
our  schools.  Yours,  &c.,  J.  D.  PHIL  BRICK. 

MAYHEW  SCHOOL. 

GRAMMATICAL  CHART  OF  SENTENCES,  by  Samuel  S.  Greene,  A.  M., 
Principal  of  Phillips  School,  Boston. 

The  above  7hart  contains  a  classification  and  illustration  of  all  the  component  parts 
of  everv  sentence,  in  the  English  language,  with  definitions,  directions  a.v.A  models,  to 
aid  th_  teacher  in  using  it.  It  is  designed  to  assist  in  teaching  the  grammar  of  the 
.anguage  on  a  more  philosophical  plan;  it  exhibits  the  structure  of  sentences,  by 
commencing  with  the  simplest  forms,  and  advancing  gradually  through  every  variety 
of  construction,  and  will  he  found  a  very  valuable  help  in  the  tedious  process  of 
teaching  grammar. 


In  Press,  and  will  shortly  be  published, 

A  GRAMMAR.  ON  AN  ENTIRELY  NEW  PLAN.  By  S.  S.  Greene,  A.  M., 
Principal  of  the  Phillips  School,  Boston,  which  is  intended  to  accompany  the  above 
Chart.  The  publishers  feel  warranted  in  saying,  (from  the  opinions  of  our  best  teach 
ers  who  have  seen  the  proof-sheets,)  that  the  above  work  will  need  only  to  be  seen  to 
be  universally  adopted ;  —  that  it  will  open  to  tho  pupil  new  light  on  this  hitherto  dry 
study. 

FOLSOM'S  LIVY.  Titi  Livii  Patavina  Historiarum  Liber  Primus  et  Selecta 
quaedam  Capita.  Curavit  Notulisque  instruxit,  CAROLUS  FOLSOM,  Academies  Har- 
vardianse  olim  Bibliothecariua.  15th  Stereotype  Edition. 

POLLEN'S  PRACTICAL  GRAMMAR  OF  THE  GERMAN  LANGUAGE.  12th 
edition. 

FOLLEN'S  GERMAN  READER,  fur  Beginners,    llth  edition. 

Follen's  German  Grammar  and  Reader  have  been  very  highly  recommended.  The 
second  and  third  editions  of  the  Grammar  W3re  both  greatly  improved,  by  corrections 
and  additional  rules  and  illustrations,  by  the  author;  since  which,  the  work  has 
passed  through  several  editions,  and  has  been  introduced  into  Harvard  University. 

The  German  Reader,  also,  has  been  repeatedly  printed ;  and  its  popularity  con 
stantly  increasing. 

A  NATURAL  HISTORY  of  the  most  remarkable  Quadrupeds,  Birds,  Fishes, 
Serpents,  Reptiles,  and  Insects.  By  Mrs.  Mary  Trimmer.  With  200  Engravings. 
Abridged  and  improved ;  particularly  designed  for  youth  of  the  United  States,  and 
suited  to  the  use  of  Schools.  19th  edition. 

MEADOWS'  FRENCH  AND  ENGLISH  PRONOUNCING  DICTIONARY.  A 

new  French  and  English  Pronouncing  Dictionary,  on  the  basis  of  Nugent's,  with 
many  new  words  in  general  use.  In  two  parts.  First,  French  and  English ;  second, 
English  and  French,  exhibiting  the  pronunciation  of  the  French  in  pure  English 
sounds  ;  the  parts  of  speech,  gender  of  French  nouns,  regular  and  irregular  conjuga 
tions  of  verbs,  accent  of  English  worda,  list  of  the  usual  Christian  and  proper  names, 
and  names  of  countries  and  nations,  to  which  are  prefixed,  principles  of  French  pro 
nunciation,  and  an  abridged  grammar.  By  F.  C.  Meadows,  M.  A.,  of  the  University 
of  Paris.  New  edition,  revised  and  improved  by  Charles  L.  Parmentier,  A.  M.,  Pro 
fessor  of  the  French  Language  and  Literature.  18mo.,  sheep. 

LE  BRUN'S  TELEMAQUE.  Les  Aventures  de  Tetemaque,  fila  d'Ulysse.  Par 
M.  Fen&on.  Edited  by  Mr.  Charles  Le  Brun.  1  vol.  12mo.,  sheep. 

THE  COMMON  SCHOOL  SONG-BOOK.  By  Asa  Fitz,  author  of  the  American 
School  Song  Book,  Primary  Song  Book,  &c. 

THE  SABBATH  SCHOOL  MINSTREL.  By  Asa  Fitz,  author  of  the  Common 
School  Sons-Book,  &c..  being  a  collection  of  pleasant  Hymns  set  to  Music,  and 
adapted  to  the  use  of  Sabbath  Schools. 

MODERN  GEOGRAPHY  AND  ATLAS.    By  J.  E.  Worcester ;  new  edition. 

ANCIENT  CLASSICAL  AND  SCRIPTURAL  GEOGRAPHY  AND  ATLAS. 
By  J.  E.  Worcester. 

ELEMENTS  OF  PLANE  GEOMETRY.  By  N.  R  Tillinghast,  Principal  of  the 
Bridgewater  Normal  School. 

VALUABLE  BOOKS  FOR  LIBRARIES. 

LIVES  OF  THE  HEROES  OF  THE  AMERICAN  REVOLUTION,  comprising 
the  Life  of  Washington  and  his  Generals  and  OlRcers  who  were  most  distinguished  in 
the  War  of  the  Independence  of  the  United  States  of  America.  Also,  the  Constitution 
of  the  United  States  and  Amendments,  the  Declaration  of  Independence,  with  Wash 
ington's  Inaugural,  First  Annual,  and  Farewell  Addresses.  Embellished  with  Por 
traits.  1  vol.,  12mo. 

LIFE  AND  CAMPAIGNS  OF  NAPOLEON  BONAPARTE,  giving  an  account 
of  all  his  engagements  from  the  siege  of  Toulon  to  the  battle  of  Waterloo.  Also,  em 
bracing  accounts  of  the  darin?  exploits  of  his  Marshals,  together  with  his  public  and 
private  life,  from  the  commencement  of  his  career  to  his  final  imprisonment  and 
death  on  St.  Helena,  translated  from  the  most  authentic  sources.  Illustrated  with 
fine  engravings.  I  vol..  12mo. 

BANCROFT'S  LIFE  OF  GEORGE  WASHINGTON,  being  the  most  impartial 
and  authentic  biography  of  this  illustrious  patriot  thnt  has  been  heretofore  published ; 
also,  embracing  a  History  of  the  Wars  of  the  American  Revolution.  I  vol.,  12mo., 
Illustrated. 

£3=  PHILLIPS  &  SAMPSON  keep  constantly  on  hand  one  of  the  largest  assort 
ments  of  SCHOOL,  CLASSICAL  AND  MISCELLANEOUS  BOOKS,  STATIONERY, 
<fcc.,  to  be  found  in  New  England,  which  will  be  sold  at  prices  as  low  as  at  any  other 
bookstore  in  the  country. ^  ^ 

48 


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